Mixing
Does $sum_{n=1}^infty frac{cos{(sqrt{n})}}{n}$ converge?
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The series is:
$$sum_{n=1}^infty frac{cos(sqrt{n})}{n}$$
Considering it isn't always positive, I replace $frac{cos{sqrt{n}}}{n}$ with its absolute value and I find that:
$$vert frac{cos{sqrt{n}}}{n}vertgt frac{cos^2{sqrt{n}}}{n}=frac{ 1+cos{2sqrt{n}}}{2n}=frac{1}{2n}+frac{cos{2sqrt{n}}}{2n}$$
if $sum_{n=1}^infty vertfrac{cos{sqrt{n}}}{n}vert $ converges, then
$sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that $sum_{n=1}^inftyfrac{1}{2n}$ converges , which is impossible.
So I get that $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ absolutely diverges. But I can't figure out whether $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges or not.
I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=sum_{k=1}^n cos{sqrt{k}}$$ is bounded.
(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)
sequences-and-series analysis trigonometry
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
$endgroup$
add a comment |
$begingroup$
The series is:
$$sum_{n=1}^infty frac{cos(sqrt{n})}{n}$$
Considering it isn't always positive, I replace $frac{cos{sqrt{n}}}{n}$ with its absolute value and I find that:
$$vert frac{cos{sqrt{n}}}{n}vertgt frac{cos^2{sqrt{n}}}{n}=frac{ 1+cos{2sqrt{n}}}{2n}=frac{1}{2n}+frac{cos{2sqrt{n}}}{2n}$$
if $sum_{n=1}^infty vertfrac{cos{sqrt{n}}}{n}vert $ converges, then
$sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that $sum_{n=1}^inftyfrac{1}{2n}$ converges , which is impossible.
So I get that $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ absolutely diverges. But I can't figure out whether $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges or not.
I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=sum_{k=1}^n cos{sqrt{k}}$$ is bounded.
(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)
sequences-and-series analysis trigonometry
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
$endgroup$
$begingroup$
If you approximate the sum by an integral, then you have $$sum_{n=1}^m,frac{cos(sqrt{n})}{n}approx int_1^m,frac{cos(sqrt{x})}{x},text{d}x=2,text{Ci}(sqrt{m})-2,text{Ci}(1),,$$ where $text{Ci}$ is the cosine integral $$text{Ci}(t)=-int_t^infty,frac{cos(s)}{s},text{d}s,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$lim_{mtoinfty},text{Ci}(sqrt{m})=0,.$$
$endgroup$
– Batominovski
Oct 9 '18 at 11:27
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Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$.
$endgroup$
– Batominovski
Oct 9 '18 at 11:32
add a comment |
$begingroup$
The series is:
$$sum_{n=1}^infty frac{cos(sqrt{n})}{n}$$
Considering it isn't always positive, I replace $frac{cos{sqrt{n}}}{n}$ with its absolute value and I find that:
$$vert frac{cos{sqrt{n}}}{n}vertgt frac{cos^2{sqrt{n}}}{n}=frac{ 1+cos{2sqrt{n}}}{2n}=frac{1}{2n}+frac{cos{2sqrt{n}}}{2n}$$
if $sum_{n=1}^infty vertfrac{cos{sqrt{n}}}{n}vert $ converges, then
$sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that $sum_{n=1}^inftyfrac{1}{2n}$ converges , which is impossible.
So I get that $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ absolutely diverges. But I can't figure out whether $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges or not.
I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=sum_{k=1}^n cos{sqrt{k}}$$ is bounded.
(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)
sequences-and-series analysis trigonometry
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
$endgroup$
The series is:
$$sum_{n=1}^infty frac{cos(sqrt{n})}{n}$$
Considering it isn't always positive, I replace $frac{cos{sqrt{n}}}{n}$ with its absolute value and I find that:
$$vert frac{cos{sqrt{n}}}{n}vertgt frac{cos^2{sqrt{n}}}{n}=frac{ 1+cos{2sqrt{n}}}{2n}=frac{1}{2n}+frac{cos{2sqrt{n}}}{2n}$$
if $sum_{n=1}^infty vertfrac{cos{sqrt{n}}}{n}vert $ converges, then
$sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that $sum_{n=1}^inftyfrac{1}{2n}$ converges , which is impossible.
So I get that $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ absolutely diverges. But I can't figure out whether $sum_{n=1}^infty frac{cos{sqrt{n}}}{n}$ converges or not.
I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=sum_{k=1}^n cos{sqrt{k}}$$ is bounded.
(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :)
sequences-and-series analysis trigonometry
sequences-and-series analysis trigonometry
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
edited Oct 9 '18 at 12:22
Robert Z
101k1071144
101k1071144
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
asked Oct 9 '18 at 10:53
hmtbgchmtbgc
1097
1097
$begingroup$
If you approximate the sum by an integral, then you have $$sum_{n=1}^m,frac{cos(sqrt{n})}{n}approx int_1^m,frac{cos(sqrt{x})}{x},text{d}x=2,text{Ci}(sqrt{m})-2,text{Ci}(1),,$$ where $text{Ci}$ is the cosine integral $$text{Ci}(t)=-int_t^infty,frac{cos(s)}{s},text{d}s,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$lim_{mtoinfty},text{Ci}(sqrt{m})=0,.$$
$endgroup$
– Batominovski
Oct 9 '18 at 11:27
$begingroup$
Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$.
$endgroup$
– Batominovski
Oct 9 '18 at 11:32
add a comment |
$begingroup$
If you approximate the sum by an integral, then you have $$sum_{n=1}^m,frac{cos(sqrt{n})}{n}approx int_1^m,frac{cos(sqrt{x})}{x},text{d}x=2,text{Ci}(sqrt{m})-2,text{Ci}(1),,$$ where $text{Ci}$ is the cosine integral $$text{Ci}(t)=-int_t^infty,frac{cos(s)}{s},text{d}s,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$lim_{mtoinfty},text{Ci}(sqrt{m})=0,.$$
$endgroup$
– Batominovski
Oct 9 '18 at 11:27
$begingroup$
Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$.
$endgroup$
– Batominovski
Oct 9 '18 at 11:32
$begingroup$
If you approximate the sum by an integral, then you have $$sum_{n=1}^m,frac{cos(sqrt{n})}{n}approx int_1^m,frac{cos(sqrt{x})}{x},text{d}x=2,text{Ci}(sqrt{m})-2,text{Ci}(1),,$$ where $text{Ci}$ is the cosine integral $$text{Ci}(t)=-int_t^infty,frac{cos(s)}{s},text{d}s,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$lim_{mtoinfty},text{Ci}(sqrt{m})=0,.$$
$endgroup$
– Batominovski
Oct 9 '18 at 11:27
$begingroup$
If you approximate the sum by an integral, then you have $$sum_{n=1}^m,frac{cos(sqrt{n})}{n}approx int_1^m,frac{cos(sqrt{x})}{x},text{d}x=2,text{Ci}(sqrt{m})-2,text{Ci}(1),,$$ where $text{Ci}$ is the cosine integral $$text{Ci}(t)=-int_t^infty,frac{cos(s)}{s},text{d}s,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$lim_{mtoinfty},text{Ci}(sqrt{m})=0,.$$
$endgroup$
– Batominovski
Oct 9 '18 at 11:27
$begingroup$
Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$.
$endgroup$
– Batominovski
Oct 9 '18 at 11:32
$begingroup$
Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$.
$endgroup$
– Batominovski
Oct 9 '18 at 11:32
add a comment |
5 Answers
5
active
oldest
votes
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We show that $Ntosum_{n = 1}^N {frac{cos(sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $ngeq 1$ and for all $xin [n,n+1)$
$$left|frac{cos(sqrt{n})}{n}-frac{cos(sqrt{x})}{x}right|leq frac{1}{n^{3/2}}.$$
Then for $M>Ngeq 1$,
$$begin{align}
&left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - int_N^{M + 1} {frac{{cos (sqrt{x} )}}{x}dx} right|\
&=left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - sum_{n = N}^M {int_n^{n + 1} {frac{{cos (sqrt{x} )}}{x}dx} } right|\
&leqsum_{n = N}^M int_n^{n + 1}left|{frac{cos(sqrt{n})}{n}} - { {frac{{cos (sqrt{x} )}}{x}} } right|dx le sum_{n = N}^M frac{1}{n^{3/2}}.end{align}$$
Note that since $sum_{n=1}^{infty}frac{1}{n^{3/2}}$ is convergent then
$$lim_{M,Nto +infty} sum_{n = N}^M frac{1}{n^{3/2}}=0.$$
Moreover
$$int_1^{+infty}frac{{cos (sqrt{x} )}}{x}dx=2int_1^{+infty}frac{{cos (u )}}{u}, du$$
where the last integral is convergent, and we have that
$$lim_{M,Nto +infty}int_N^{M+1}frac{{cos (sqrt{x} )}}{x}dx=0.$$
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
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So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
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– Parcly Taxel
Oct 9 '18 at 11:49
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Not sure how the first inequality can lead to the second one.
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– Szeto
Oct 9 '18 at 11:50
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@Szeto Is it clear now?
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– Robert Z
Oct 9 '18 at 11:55
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Thanks for your excellent proof!
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– hmtbgc
Oct 9 '18 at 12:55
add a comment |
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Not a theoretical answer but a computational one. Ran a SageMath code to check the sum. It appears that the sum is bounded between oscillates in the neighbourhood of $1/3$.
n = 1
s = 0
s_min = s_max = -0.33
while(n < 10^12):
s = s + cos(n^0.5)/n
if(s < s_min):
s_min = s
if(s > s_max):
s_max = s
if(n%10^6 == 0):
print(n,s,s_min,s_max)
n = n + 1
Given below is the sum of the first 540 million terms. I will keep updating the sum after every 100 million terms to see if and where it shows hints of convergence. Looking at these numbers, I wonder if the sum oscillates between -0.3307 and 0.3306.
n s_n
(537000000, -0.330621546932186)
(538000000, -0.330725062788227)
(539000000, -0.330687353946068)
(540000000, -0.330649408748269)
(541000000, -0.330757389552252)
(542000000, -0.330602401857676)
(543000000, -0.330766627976051)
(544000000, -0.330637600125639)
(545000000, -0.330692819220692)
(546000000, -0.330730454234348)
(547000000, -0.330610479504947)
(548000000, -0.330771768297931)
(549000000, -0.330629435680553)
(550000000, -0.330695312033157)
(551000000, -0.330735165460147)
(552000000, -0.330605859570737)
(553000000, -0.330764635324360)
(554000000, -0.330655186664157)
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
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A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
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– Winther
Oct 9 '18 at 12:01
1
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Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
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– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
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Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
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– Mariusz Iwaniuk
Oct 24 '18 at 17:42
add a comment |
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Notations: $lfloor x rfloor$ is the floor function, ${x}$ is the fractional part of $x$ so that $x=lfloor xrfloor + {x}$.
Applying partial summation with $f(x)=frac{cos(sqrt x)}x$,
$$begin{align}
sum_{n=1}^N frac{cos(sqrt n)}n&=int_{1-}^N f(x)dlfloor x rfloor \
&=f(x)lfloor xrfloor Bigvert_{1-}^N-int_{1-}^N f'(x)lfloor xrfloor dx\
&=f(N)(N-{N})-int_1^N xf'(x)dx+int_1^N{x}f'(x)dx\
&=Nf(N)-f(1)-int_1^Nxf'(x)dx+int_1^N{x}f'(x)dx+f(1)-{N}f(N).
end{align}
$$
From integration by parts, the sum of first three terms is
$$
int_1^{N}f(x)dx
$$
Thus, we have the following as $Nrightarrowinfty$,
$$
sum_{n=1}^{infty}frac{cos(sqrt n)}n=int_1^{infty}f(x)dx+int_1^{infty}{x}f'(x)dx+cos 1.
$$
It is easy to see that the integrals converge.
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
$endgroup$
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
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– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
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$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
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– i707107
Oct 11 '18 at 2:13
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Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
add a comment |
$begingroup$
This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $d lfloor x rfloor$.
Letting
$$c_{k} = c(k) = frac{cos(sqrt{k})}{k}$$
we attempt to find an integral representation for the partial sum
$$s_n = sum_{k=1}^n c_{k}$$
The formulas for partial summation are
$$sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$
$$A_{k} = sum_{i=1}^k a_{i}$$
Letting $a_{k}=1, b_{k} = c_{k}$ we have $A_{k} = k$.
Now comes the trick which introduces an integral: we have
$$(b_{k}-b_{k+1}) = - int_{k}^{k+1} c'(x);dx$$
and, what's more, the factor $k$ can be incorporated in the integral:
$$k (b_{k}-b_{k+1}) = - k int_{k}^{k+1} c'(x);dx = - int_{k}^{k+1} lfloor xrfloor c'(x);dx $$
Where $lfloor xrfloor$ is the floor function.
Hence, using
$$lfloor xrfloor = x -{x} $$
where ${x}$ is the fractional part of $x$, the partial sum becomes
$$s_{n} = n c_n -sum_{k=1}^{n-1} int_{k}^{k+1} lfloor xrfloor c'(x);dx \
= n c_n -int_{1}^{n} x c'(x);dx + int_{1}^{n} {x} c'(x);dx$$
By partial integration of the first integral the term $n c_{n}$ drops out and we get
$$s_{n} = cos(1) +int_{1}^{n} c(x);dx + int_{1}^{n} {x} c'(x);dx$$
The first integral can be solved explicitly
$$int_{1}^{n} c(x);dx = 2 text{Ci}left(sqrt{n}right)-2 text{Ci}(1)$$
Where $text{Ci}$ is the integral cosine.
The second integral is absolutely convergent and can be very roughly estimated thus (notice that $0le {x} lt 1$)
$$|i_2| = |int_{1}^{n} {x} c'(x),dx| <= int_{1}^{n}| {x} c'(x)|,dx \
<= int_{1}^{n} |{x}| |c'(x)|,dx
<=int_{1}^{n} | c'(x) | ,dx$$
Now
$$| c'(x) | = |frac{cos left(sqrt{x}right)}{x^2}+frac{sin left(sqrt{x}right)}{2 x^{3/2}}| \
leq | frac{sin left(sqrt{x}right)}{2 x^{3/2}}| +|frac{cos left(sqrt{x}right)}{x^2}| leq frac{1}{2 x^{3/2}}+frac{1}{x^2}$$
Hence
$$| i_2| <=int_1^n left(frac{1}{2 x^{3/2}}+frac{1}{x^2}right) , dx = 2-frac{sqrt{n}+1}{n}tag{*}$$
Since $lim_{nto infty } , text{Ci}left(sqrt{n}right)= 0$ the limit of the partial sum is given by
$$s = cos(1) - 2 text{Ci}(1) + lim_{nto infty } ,i_2$$
Observing (*) $s$ remains finite as $ntoinfty$ hence the original sum is convergent. QED.
Remark
A more sophisticated study if the integral $i_2$ might provide better numerical bounds, and even a closed form.
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
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add a comment |
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This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.
Step 1: Transformation to an alternating sum
The partial sum in question is
$$s_n = sum_{k=1}^n c_ktag{1}$$
where
$$c_k = frac{cos(sqrt{k})}{k}tag{2}$$
Now collecting all subsequent summands with the same sign we can write
$$s_n = sum_{m=0}^M (-1)^m f_mtag{3}$$
where
$$f_0 = sum_{k=1}^{lfloor z_{1}rfloor} c_{k}tag{4a}$$
$$f_{mge 1} = (-1)^m sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} c_{k}tag{4b}$$
is the sum of terms between two adjacent roots $k=z_m$ and $k=z_{m+1}$ of $c(k) = 0$, and $M$ is some number depending on $n$ which can in principle be specified but it is normally not necessary as it goes to $infty$ together with $n$
Then, by an appropirate choice of the factor $(-1)^m$, all $f_m$ can be chosen to be positive quantities.
Step 2: Application of Dirichlet's criterion
Now we can apply the Dirichlet criterion to (3). This reads now as follows: $s_n$ is convergent if
(1) $f_m to 0$ for $mtoinfty$
(2) $f_m$ is monotonous
ad (1)
We have for $mge1$
$$|f_{m}| le g(m)$$
where we have dropped the $cos$ and have defined
$$g(m) = sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} frac{1}{k}
=H_{lfloor z_{m+1} rfloor}-H_{lceil z_{m}rceil-1}tag{5} $$
Here $H_n$ ist the harmonic number.
Now the zeroes of $c_{k}$ are given by
$$z_m = left(pi(m-frac{1}{2})right)^2tag{6}$$
Dropping Floor and Ceiling in $g(m)$ defines
$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$
Letting ${lfloor x rfloor} to x-1,{lceil x rceil} to x+1 $ defines another function
$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$
and we have the inequality
$$g_1(m) lt g(m) lt g_0(m)$$
Asymptotically this leads, up to $O(frac{1}{m^2})$, to the inequality
$$frac{2}{m}- frac{1}{pi^2 m^2}< g(m) <frac{2}{m}+frac{1}{pi^2 m^2} $$
This shows that $f_m$ goes to zero.
ad (2)
(to be continued)
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
add a comment |
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5 Answers
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5 Answers
5
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active
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active
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$begingroup$
We show that $Ntosum_{n = 1}^N {frac{cos(sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $ngeq 1$ and for all $xin [n,n+1)$
$$left|frac{cos(sqrt{n})}{n}-frac{cos(sqrt{x})}{x}right|leq frac{1}{n^{3/2}}.$$
Then for $M>Ngeq 1$,
$$begin{align}
&left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - int_N^{M + 1} {frac{{cos (sqrt{x} )}}{x}dx} right|\
&=left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - sum_{n = N}^M {int_n^{n + 1} {frac{{cos (sqrt{x} )}}{x}dx} } right|\
&leqsum_{n = N}^M int_n^{n + 1}left|{frac{cos(sqrt{n})}{n}} - { {frac{{cos (sqrt{x} )}}{x}} } right|dx le sum_{n = N}^M frac{1}{n^{3/2}}.end{align}$$
Note that since $sum_{n=1}^{infty}frac{1}{n^{3/2}}$ is convergent then
$$lim_{M,Nto +infty} sum_{n = N}^M frac{1}{n^{3/2}}=0.$$
Moreover
$$int_1^{+infty}frac{{cos (sqrt{x} )}}{x}dx=2int_1^{+infty}frac{{cos (u )}}{u}, du$$
where the last integral is convergent, and we have that
$$lim_{M,Nto +infty}int_N^{M+1}frac{{cos (sqrt{x} )}}{x}dx=0.$$
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
$endgroup$
$begingroup$
So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
$endgroup$
– Parcly Taxel
Oct 9 '18 at 11:49
$begingroup$
Not sure how the first inequality can lead to the second one.
$endgroup$
– Szeto
Oct 9 '18 at 11:50
$begingroup$
@Szeto Is it clear now?
$endgroup$
– Robert Z
Oct 9 '18 at 11:55
$begingroup$
Thanks for your excellent proof!
$endgroup$
– hmtbgc
Oct 9 '18 at 12:55
add a comment |
$begingroup$
We show that $Ntosum_{n = 1}^N {frac{cos(sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $ngeq 1$ and for all $xin [n,n+1)$
$$left|frac{cos(sqrt{n})}{n}-frac{cos(sqrt{x})}{x}right|leq frac{1}{n^{3/2}}.$$
Then for $M>Ngeq 1$,
$$begin{align}
&left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - int_N^{M + 1} {frac{{cos (sqrt{x} )}}{x}dx} right|\
&=left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - sum_{n = N}^M {int_n^{n + 1} {frac{{cos (sqrt{x} )}}{x}dx} } right|\
&leqsum_{n = N}^M int_n^{n + 1}left|{frac{cos(sqrt{n})}{n}} - { {frac{{cos (sqrt{x} )}}{x}} } right|dx le sum_{n = N}^M frac{1}{n^{3/2}}.end{align}$$
Note that since $sum_{n=1}^{infty}frac{1}{n^{3/2}}$ is convergent then
$$lim_{M,Nto +infty} sum_{n = N}^M frac{1}{n^{3/2}}=0.$$
Moreover
$$int_1^{+infty}frac{{cos (sqrt{x} )}}{x}dx=2int_1^{+infty}frac{{cos (u )}}{u}, du$$
where the last integral is convergent, and we have that
$$lim_{M,Nto +infty}int_N^{M+1}frac{{cos (sqrt{x} )}}{x}dx=0.$$
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
$endgroup$
$begingroup$
So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
$endgroup$
– Parcly Taxel
Oct 9 '18 at 11:49
$begingroup$
Not sure how the first inequality can lead to the second one.
$endgroup$
– Szeto
Oct 9 '18 at 11:50
$begingroup$
@Szeto Is it clear now?
$endgroup$
– Robert Z
Oct 9 '18 at 11:55
$begingroup$
Thanks for your excellent proof!
$endgroup$
– hmtbgc
Oct 9 '18 at 12:55
add a comment |
$begingroup$
We show that $Ntosum_{n = 1}^N {frac{cos(sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $ngeq 1$ and for all $xin [n,n+1)$
$$left|frac{cos(sqrt{n})}{n}-frac{cos(sqrt{x})}{x}right|leq frac{1}{n^{3/2}}.$$
Then for $M>Ngeq 1$,
$$begin{align}
&left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - int_N^{M + 1} {frac{{cos (sqrt{x} )}}{x}dx} right|\
&=left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - sum_{n = N}^M {int_n^{n + 1} {frac{{cos (sqrt{x} )}}{x}dx} } right|\
&leqsum_{n = N}^M int_n^{n + 1}left|{frac{cos(sqrt{n})}{n}} - { {frac{{cos (sqrt{x} )}}{x}} } right|dx le sum_{n = N}^M frac{1}{n^{3/2}}.end{align}$$
Note that since $sum_{n=1}^{infty}frac{1}{n^{3/2}}$ is convergent then
$$lim_{M,Nto +infty} sum_{n = N}^M frac{1}{n^{3/2}}=0.$$
Moreover
$$int_1^{+infty}frac{{cos (sqrt{x} )}}{x}dx=2int_1^{+infty}frac{{cos (u )}}{u}, du$$
where the last integral is convergent, and we have that
$$lim_{M,Nto +infty}int_N^{M+1}frac{{cos (sqrt{x} )}}{x}dx=0.$$
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
$endgroup$
We show that $Ntosum_{n = 1}^N {frac{cos(sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $ngeq 1$ and for all $xin [n,n+1)$
$$left|frac{cos(sqrt{n})}{n}-frac{cos(sqrt{x})}{x}right|leq frac{1}{n^{3/2}}.$$
Then for $M>Ngeq 1$,
$$begin{align}
&left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - int_N^{M + 1} {frac{{cos (sqrt{x} )}}{x}dx} right|\
&=left|sum_{n = N}^M {frac{cos(sqrt{n})}{n}} - sum_{n = N}^M {int_n^{n + 1} {frac{{cos (sqrt{x} )}}{x}dx} } right|\
&leqsum_{n = N}^M int_n^{n + 1}left|{frac{cos(sqrt{n})}{n}} - { {frac{{cos (sqrt{x} )}}{x}} } right|dx le sum_{n = N}^M frac{1}{n^{3/2}}.end{align}$$
Note that since $sum_{n=1}^{infty}frac{1}{n^{3/2}}$ is convergent then
$$lim_{M,Nto +infty} sum_{n = N}^M frac{1}{n^{3/2}}=0.$$
Moreover
$$int_1^{+infty}frac{{cos (sqrt{x} )}}{x}dx=2int_1^{+infty}frac{{cos (u )}}{u}, du$$
where the last integral is convergent, and we have that
$$lim_{M,Nto +infty}int_N^{M+1}frac{{cos (sqrt{x} )}}{x}dx=0.$$
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
edited Oct 9 '18 at 13:08
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
answered Oct 9 '18 at 11:46
Robert ZRobert Z
101k1071144
101k1071144
$begingroup$
So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
$endgroup$
– Parcly Taxel
Oct 9 '18 at 11:49
$begingroup$
Not sure how the first inequality can lead to the second one.
$endgroup$
– Szeto
Oct 9 '18 at 11:50
$begingroup$
@Szeto Is it clear now?
$endgroup$
– Robert Z
Oct 9 '18 at 11:55
$begingroup$
Thanks for your excellent proof!
$endgroup$
– hmtbgc
Oct 9 '18 at 12:55
add a comment |
$begingroup$
So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
$endgroup$
– Parcly Taxel
Oct 9 '18 at 11:49
$begingroup$
Not sure how the first inequality can lead to the second one.
$endgroup$
– Szeto
Oct 9 '18 at 11:50
$begingroup$
@Szeto Is it clear now?
$endgroup$
– Robert Z
Oct 9 '18 at 11:55
$begingroup$
Thanks for your excellent proof!
$endgroup$
– hmtbgc
Oct 9 '18 at 12:55
$begingroup$
So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
$endgroup$
– Parcly Taxel
Oct 9 '18 at 11:49
$begingroup$
So ultimately the absolute-term series is less than $zeta(3/2)$ and thus the original series converges?
$endgroup$
– Parcly Taxel
Oct 9 '18 at 11:49
$begingroup$
Not sure how the first inequality can lead to the second one.
$endgroup$
– Szeto
Oct 9 '18 at 11:50
$begingroup$
Not sure how the first inequality can lead to the second one.
$endgroup$
– Szeto
Oct 9 '18 at 11:50
$begingroup$
@Szeto Is it clear now?
$endgroup$
– Robert Z
Oct 9 '18 at 11:55
$begingroup$
@Szeto Is it clear now?
$endgroup$
– Robert Z
Oct 9 '18 at 11:55
$begingroup$
Thanks for your excellent proof!
$endgroup$
– hmtbgc
Oct 9 '18 at 12:55
$begingroup$
Thanks for your excellent proof!
$endgroup$
– hmtbgc
Oct 9 '18 at 12:55
add a comment |
$begingroup$
Not a theoretical answer but a computational one. Ran a SageMath code to check the sum. It appears that the sum is bounded between oscillates in the neighbourhood of $1/3$.
n = 1
s = 0
s_min = s_max = -0.33
while(n < 10^12):
s = s + cos(n^0.5)/n
if(s < s_min):
s_min = s
if(s > s_max):
s_max = s
if(n%10^6 == 0):
print(n,s,s_min,s_max)
n = n + 1
Given below is the sum of the first 540 million terms. I will keep updating the sum after every 100 million terms to see if and where it shows hints of convergence. Looking at these numbers, I wonder if the sum oscillates between -0.3307 and 0.3306.
n s_n
(537000000, -0.330621546932186)
(538000000, -0.330725062788227)
(539000000, -0.330687353946068)
(540000000, -0.330649408748269)
(541000000, -0.330757389552252)
(542000000, -0.330602401857676)
(543000000, -0.330766627976051)
(544000000, -0.330637600125639)
(545000000, -0.330692819220692)
(546000000, -0.330730454234348)
(547000000, -0.330610479504947)
(548000000, -0.330771768297931)
(549000000, -0.330629435680553)
(550000000, -0.330695312033157)
(551000000, -0.330735165460147)
(552000000, -0.330605859570737)
(553000000, -0.330764635324360)
(554000000, -0.330655186664157)
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
$endgroup$
$begingroup$
A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
$endgroup$
– Winther
Oct 9 '18 at 12:01
1
$begingroup$
Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
$endgroup$
– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
$begingroup$
Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
$endgroup$
– Mariusz Iwaniuk
Oct 24 '18 at 17:42
add a comment |
$begingroup$
Not a theoretical answer but a computational one. Ran a SageMath code to check the sum. It appears that the sum is bounded between oscillates in the neighbourhood of $1/3$.
n = 1
s = 0
s_min = s_max = -0.33
while(n < 10^12):
s = s + cos(n^0.5)/n
if(s < s_min):
s_min = s
if(s > s_max):
s_max = s
if(n%10^6 == 0):
print(n,s,s_min,s_max)
n = n + 1
Given below is the sum of the first 540 million terms. I will keep updating the sum after every 100 million terms to see if and where it shows hints of convergence. Looking at these numbers, I wonder if the sum oscillates between -0.3307 and 0.3306.
n s_n
(537000000, -0.330621546932186)
(538000000, -0.330725062788227)
(539000000, -0.330687353946068)
(540000000, -0.330649408748269)
(541000000, -0.330757389552252)
(542000000, -0.330602401857676)
(543000000, -0.330766627976051)
(544000000, -0.330637600125639)
(545000000, -0.330692819220692)
(546000000, -0.330730454234348)
(547000000, -0.330610479504947)
(548000000, -0.330771768297931)
(549000000, -0.330629435680553)
(550000000, -0.330695312033157)
(551000000, -0.330735165460147)
(552000000, -0.330605859570737)
(553000000, -0.330764635324360)
(554000000, -0.330655186664157)
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
$endgroup$
$begingroup$
A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
$endgroup$
– Winther
Oct 9 '18 at 12:01
1
$begingroup$
Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
$endgroup$
– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
$begingroup$
Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
$endgroup$
– Mariusz Iwaniuk
Oct 24 '18 at 17:42
add a comment |
$begingroup$
Not a theoretical answer but a computational one. Ran a SageMath code to check the sum. It appears that the sum is bounded between oscillates in the neighbourhood of $1/3$.
n = 1
s = 0
s_min = s_max = -0.33
while(n < 10^12):
s = s + cos(n^0.5)/n
if(s < s_min):
s_min = s
if(s > s_max):
s_max = s
if(n%10^6 == 0):
print(n,s,s_min,s_max)
n = n + 1
Given below is the sum of the first 540 million terms. I will keep updating the sum after every 100 million terms to see if and where it shows hints of convergence. Looking at these numbers, I wonder if the sum oscillates between -0.3307 and 0.3306.
n s_n
(537000000, -0.330621546932186)
(538000000, -0.330725062788227)
(539000000, -0.330687353946068)
(540000000, -0.330649408748269)
(541000000, -0.330757389552252)
(542000000, -0.330602401857676)
(543000000, -0.330766627976051)
(544000000, -0.330637600125639)
(545000000, -0.330692819220692)
(546000000, -0.330730454234348)
(547000000, -0.330610479504947)
(548000000, -0.330771768297931)
(549000000, -0.330629435680553)
(550000000, -0.330695312033157)
(551000000, -0.330735165460147)
(552000000, -0.330605859570737)
(553000000, -0.330764635324360)
(554000000, -0.330655186664157)
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
$endgroup$
Not a theoretical answer but a computational one. Ran a SageMath code to check the sum. It appears that the sum is bounded between oscillates in the neighbourhood of $1/3$.
n = 1
s = 0
s_min = s_max = -0.33
while(n < 10^12):
s = s + cos(n^0.5)/n
if(s < s_min):
s_min = s
if(s > s_max):
s_max = s
if(n%10^6 == 0):
print(n,s,s_min,s_max)
n = n + 1
Given below is the sum of the first 540 million terms. I will keep updating the sum after every 100 million terms to see if and where it shows hints of convergence. Looking at these numbers, I wonder if the sum oscillates between -0.3307 and 0.3306.
n s_n
(537000000, -0.330621546932186)
(538000000, -0.330725062788227)
(539000000, -0.330687353946068)
(540000000, -0.330649408748269)
(541000000, -0.330757389552252)
(542000000, -0.330602401857676)
(543000000, -0.330766627976051)
(544000000, -0.330637600125639)
(545000000, -0.330692819220692)
(546000000, -0.330730454234348)
(547000000, -0.330610479504947)
(548000000, -0.330771768297931)
(549000000, -0.330629435680553)
(550000000, -0.330695312033157)
(551000000, -0.330735165460147)
(552000000, -0.330605859570737)
(553000000, -0.330764635324360)
(554000000, -0.330655186664157)
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
edited Oct 9 '18 at 15:37
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
answered Oct 9 '18 at 11:40
Nilotpal Kanti SinhaNilotpal Kanti Sinha
4,43021640
4,43021640
$begingroup$
A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
$endgroup$
– Winther
Oct 9 '18 at 12:01
1
$begingroup$
Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
$endgroup$
– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
$begingroup$
Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
$endgroup$
– Mariusz Iwaniuk
Oct 24 '18 at 17:42
add a comment |
$begingroup$
A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
$endgroup$
– Winther
Oct 9 '18 at 12:01
1
$begingroup$
Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
$endgroup$
– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
$begingroup$
Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
$endgroup$
– Mariusz Iwaniuk
Oct 24 '18 at 17:42
$begingroup$
A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
$endgroup$
– Winther
Oct 9 '18 at 12:01
$begingroup$
A quick estimate suggest you roughly need $sim 10^{2k}$ terms to get convergence to $k$ digits.
$endgroup$
– Winther
Oct 9 '18 at 12:01
1
1
$begingroup$
Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
$endgroup$
– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
$begingroup$
Letting $a(k) = frac{cos left(sqrt{k}right)}{k}$ Mathematica gives $s=text{NSum[a[k], {k, 1, infty}]} = -0.330688$
$endgroup$
– Dr. Wolfgang Hintze
Oct 9 '18 at 14:10
$begingroup$
Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
$endgroup$
– Mariusz Iwaniuk
Oct 24 '18 at 17:42
$begingroup$
Numeric inverse Laplace: $gamma +mathcal{L}_s^{-1}left[frac{psi left(1+s^2right)}{s}right](1)approx -0.33068768045214513116891754615748$ where $psi$ is PolyGamma and $gamma$ is EulerGamma.
$endgroup$
– Mariusz Iwaniuk
Oct 24 '18 at 17:42
add a comment |
$begingroup$
Notations: $lfloor x rfloor$ is the floor function, ${x}$ is the fractional part of $x$ so that $x=lfloor xrfloor + {x}$.
Applying partial summation with $f(x)=frac{cos(sqrt x)}x$,
$$begin{align}
sum_{n=1}^N frac{cos(sqrt n)}n&=int_{1-}^N f(x)dlfloor x rfloor \
&=f(x)lfloor xrfloor Bigvert_{1-}^N-int_{1-}^N f'(x)lfloor xrfloor dx\
&=f(N)(N-{N})-int_1^N xf'(x)dx+int_1^N{x}f'(x)dx\
&=Nf(N)-f(1)-int_1^Nxf'(x)dx+int_1^N{x}f'(x)dx+f(1)-{N}f(N).
end{align}
$$
From integration by parts, the sum of first three terms is
$$
int_1^{N}f(x)dx
$$
Thus, we have the following as $Nrightarrowinfty$,
$$
sum_{n=1}^{infty}frac{cos(sqrt n)}n=int_1^{infty}f(x)dx+int_1^{infty}{x}f'(x)dx+cos 1.
$$
It is easy to see that the integrals converge.
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
$endgroup$
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
$endgroup$
– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
$begingroup$
$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
$endgroup$
– i707107
Oct 11 '18 at 2:13
$begingroup$
Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
add a comment |
$begingroup$
Notations: $lfloor x rfloor$ is the floor function, ${x}$ is the fractional part of $x$ so that $x=lfloor xrfloor + {x}$.
Applying partial summation with $f(x)=frac{cos(sqrt x)}x$,
$$begin{align}
sum_{n=1}^N frac{cos(sqrt n)}n&=int_{1-}^N f(x)dlfloor x rfloor \
&=f(x)lfloor xrfloor Bigvert_{1-}^N-int_{1-}^N f'(x)lfloor xrfloor dx\
&=f(N)(N-{N})-int_1^N xf'(x)dx+int_1^N{x}f'(x)dx\
&=Nf(N)-f(1)-int_1^Nxf'(x)dx+int_1^N{x}f'(x)dx+f(1)-{N}f(N).
end{align}
$$
From integration by parts, the sum of first three terms is
$$
int_1^{N}f(x)dx
$$
Thus, we have the following as $Nrightarrowinfty$,
$$
sum_{n=1}^{infty}frac{cos(sqrt n)}n=int_1^{infty}f(x)dx+int_1^{infty}{x}f'(x)dx+cos 1.
$$
It is easy to see that the integrals converge.
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
$endgroup$
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
$endgroup$
– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
$begingroup$
$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
$endgroup$
– i707107
Oct 11 '18 at 2:13
$begingroup$
Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
add a comment |
$begingroup$
Notations: $lfloor x rfloor$ is the floor function, ${x}$ is the fractional part of $x$ so that $x=lfloor xrfloor + {x}$.
Applying partial summation with $f(x)=frac{cos(sqrt x)}x$,
$$begin{align}
sum_{n=1}^N frac{cos(sqrt n)}n&=int_{1-}^N f(x)dlfloor x rfloor \
&=f(x)lfloor xrfloor Bigvert_{1-}^N-int_{1-}^N f'(x)lfloor xrfloor dx\
&=f(N)(N-{N})-int_1^N xf'(x)dx+int_1^N{x}f'(x)dx\
&=Nf(N)-f(1)-int_1^Nxf'(x)dx+int_1^N{x}f'(x)dx+f(1)-{N}f(N).
end{align}
$$
From integration by parts, the sum of first three terms is
$$
int_1^{N}f(x)dx
$$
Thus, we have the following as $Nrightarrowinfty$,
$$
sum_{n=1}^{infty}frac{cos(sqrt n)}n=int_1^{infty}f(x)dx+int_1^{infty}{x}f'(x)dx+cos 1.
$$
It is easy to see that the integrals converge.
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
$endgroup$
Notations: $lfloor x rfloor$ is the floor function, ${x}$ is the fractional part of $x$ so that $x=lfloor xrfloor + {x}$.
Applying partial summation with $f(x)=frac{cos(sqrt x)}x$,
$$begin{align}
sum_{n=1}^N frac{cos(sqrt n)}n&=int_{1-}^N f(x)dlfloor x rfloor \
&=f(x)lfloor xrfloor Bigvert_{1-}^N-int_{1-}^N f'(x)lfloor xrfloor dx\
&=f(N)(N-{N})-int_1^N xf'(x)dx+int_1^N{x}f'(x)dx\
&=Nf(N)-f(1)-int_1^Nxf'(x)dx+int_1^N{x}f'(x)dx+f(1)-{N}f(N).
end{align}
$$
From integration by parts, the sum of first three terms is
$$
int_1^{N}f(x)dx
$$
Thus, we have the following as $Nrightarrowinfty$,
$$
sum_{n=1}^{infty}frac{cos(sqrt n)}n=int_1^{infty}f(x)dx+int_1^{infty}{x}f'(x)dx+cos 1.
$$
It is easy to see that the integrals converge.
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
answered Oct 10 '18 at 5:08
i707107i707107
12.6k21647
12.6k21647
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
$endgroup$
– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
$begingroup$
$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
$endgroup$
– i707107
Oct 11 '18 at 2:13
$begingroup$
Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
add a comment |
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
$endgroup$
– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
$begingroup$
$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
$endgroup$
– i707107
Oct 11 '18 at 2:13
$begingroup$
Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
$endgroup$
– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
$begingroup$
@ i707107 I have derived the same final formula, but without using the differential $dlfloor xrfloor$ which looks strange to me. Could you please explain its meaning.
$endgroup$
– Dr. Wolfgang Hintze
Oct 10 '18 at 13:06
$begingroup$
$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
$endgroup$
– i707107
Oct 11 '18 at 2:13
$begingroup$
$dlfloor xrfloor$ means that you take the jump amount whenever there is a jump. The jumps of $lfloor x rfloor$ occur at integers with amount $1$. So, the integral on the right in fact means the same as the sum on the left. The formula I use is a partial summation formula, this is Abel's summation formula in wikipedia.
$endgroup$
– i707107
Oct 11 '18 at 2:13
$begingroup$
Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
$begingroup$
Also, you can refer to Riemann-Stieltjes integral.
$endgroup$
– i707107
Oct 11 '18 at 2:14
add a comment |
$begingroup$
This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $d lfloor x rfloor$.
Letting
$$c_{k} = c(k) = frac{cos(sqrt{k})}{k}$$
we attempt to find an integral representation for the partial sum
$$s_n = sum_{k=1}^n c_{k}$$
The formulas for partial summation are
$$sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$
$$A_{k} = sum_{i=1}^k a_{i}$$
Letting $a_{k}=1, b_{k} = c_{k}$ we have $A_{k} = k$.
Now comes the trick which introduces an integral: we have
$$(b_{k}-b_{k+1}) = - int_{k}^{k+1} c'(x);dx$$
and, what's more, the factor $k$ can be incorporated in the integral:
$$k (b_{k}-b_{k+1}) = - k int_{k}^{k+1} c'(x);dx = - int_{k}^{k+1} lfloor xrfloor c'(x);dx $$
Where $lfloor xrfloor$ is the floor function.
Hence, using
$$lfloor xrfloor = x -{x} $$
where ${x}$ is the fractional part of $x$, the partial sum becomes
$$s_{n} = n c_n -sum_{k=1}^{n-1} int_{k}^{k+1} lfloor xrfloor c'(x);dx \
= n c_n -int_{1}^{n} x c'(x);dx + int_{1}^{n} {x} c'(x);dx$$
By partial integration of the first integral the term $n c_{n}$ drops out and we get
$$s_{n} = cos(1) +int_{1}^{n} c(x);dx + int_{1}^{n} {x} c'(x);dx$$
The first integral can be solved explicitly
$$int_{1}^{n} c(x);dx = 2 text{Ci}left(sqrt{n}right)-2 text{Ci}(1)$$
Where $text{Ci}$ is the integral cosine.
The second integral is absolutely convergent and can be very roughly estimated thus (notice that $0le {x} lt 1$)
$$|i_2| = |int_{1}^{n} {x} c'(x),dx| <= int_{1}^{n}| {x} c'(x)|,dx \
<= int_{1}^{n} |{x}| |c'(x)|,dx
<=int_{1}^{n} | c'(x) | ,dx$$
Now
$$| c'(x) | = |frac{cos left(sqrt{x}right)}{x^2}+frac{sin left(sqrt{x}right)}{2 x^{3/2}}| \
leq | frac{sin left(sqrt{x}right)}{2 x^{3/2}}| +|frac{cos left(sqrt{x}right)}{x^2}| leq frac{1}{2 x^{3/2}}+frac{1}{x^2}$$
Hence
$$| i_2| <=int_1^n left(frac{1}{2 x^{3/2}}+frac{1}{x^2}right) , dx = 2-frac{sqrt{n}+1}{n}tag{*}$$
Since $lim_{nto infty } , text{Ci}left(sqrt{n}right)= 0$ the limit of the partial sum is given by
$$s = cos(1) - 2 text{Ci}(1) + lim_{nto infty } ,i_2$$
Observing (*) $s$ remains finite as $ntoinfty$ hence the original sum is convergent. QED.
Remark
A more sophisticated study if the integral $i_2$ might provide better numerical bounds, and even a closed form.
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
add a comment |
$begingroup$
This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $d lfloor x rfloor$.
Letting
$$c_{k} = c(k) = frac{cos(sqrt{k})}{k}$$
we attempt to find an integral representation for the partial sum
$$s_n = sum_{k=1}^n c_{k}$$
The formulas for partial summation are
$$sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$
$$A_{k} = sum_{i=1}^k a_{i}$$
Letting $a_{k}=1, b_{k} = c_{k}$ we have $A_{k} = k$.
Now comes the trick which introduces an integral: we have
$$(b_{k}-b_{k+1}) = - int_{k}^{k+1} c'(x);dx$$
and, what's more, the factor $k$ can be incorporated in the integral:
$$k (b_{k}-b_{k+1}) = - k int_{k}^{k+1} c'(x);dx = - int_{k}^{k+1} lfloor xrfloor c'(x);dx $$
Where $lfloor xrfloor$ is the floor function.
Hence, using
$$lfloor xrfloor = x -{x} $$
where ${x}$ is the fractional part of $x$, the partial sum becomes
$$s_{n} = n c_n -sum_{k=1}^{n-1} int_{k}^{k+1} lfloor xrfloor c'(x);dx \
= n c_n -int_{1}^{n} x c'(x);dx + int_{1}^{n} {x} c'(x);dx$$
By partial integration of the first integral the term $n c_{n}$ drops out and we get
$$s_{n} = cos(1) +int_{1}^{n} c(x);dx + int_{1}^{n} {x} c'(x);dx$$
The first integral can be solved explicitly
$$int_{1}^{n} c(x);dx = 2 text{Ci}left(sqrt{n}right)-2 text{Ci}(1)$$
Where $text{Ci}$ is the integral cosine.
The second integral is absolutely convergent and can be very roughly estimated thus (notice that $0le {x} lt 1$)
$$|i_2| = |int_{1}^{n} {x} c'(x),dx| <= int_{1}^{n}| {x} c'(x)|,dx \
<= int_{1}^{n} |{x}| |c'(x)|,dx
<=int_{1}^{n} | c'(x) | ,dx$$
Now
$$| c'(x) | = |frac{cos left(sqrt{x}right)}{x^2}+frac{sin left(sqrt{x}right)}{2 x^{3/2}}| \
leq | frac{sin left(sqrt{x}right)}{2 x^{3/2}}| +|frac{cos left(sqrt{x}right)}{x^2}| leq frac{1}{2 x^{3/2}}+frac{1}{x^2}$$
Hence
$$| i_2| <=int_1^n left(frac{1}{2 x^{3/2}}+frac{1}{x^2}right) , dx = 2-frac{sqrt{n}+1}{n}tag{*}$$
Since $lim_{nto infty } , text{Ci}left(sqrt{n}right)= 0$ the limit of the partial sum is given by
$$s = cos(1) - 2 text{Ci}(1) + lim_{nto infty } ,i_2$$
Observing (*) $s$ remains finite as $ntoinfty$ hence the original sum is convergent. QED.
Remark
A more sophisticated study if the integral $i_2$ might provide better numerical bounds, and even a closed form.
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
add a comment |
$begingroup$
This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $d lfloor x rfloor$.
Letting
$$c_{k} = c(k) = frac{cos(sqrt{k})}{k}$$
we attempt to find an integral representation for the partial sum
$$s_n = sum_{k=1}^n c_{k}$$
The formulas for partial summation are
$$sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$
$$A_{k} = sum_{i=1}^k a_{i}$$
Letting $a_{k}=1, b_{k} = c_{k}$ we have $A_{k} = k$.
Now comes the trick which introduces an integral: we have
$$(b_{k}-b_{k+1}) = - int_{k}^{k+1} c'(x);dx$$
and, what's more, the factor $k$ can be incorporated in the integral:
$$k (b_{k}-b_{k+1}) = - k int_{k}^{k+1} c'(x);dx = - int_{k}^{k+1} lfloor xrfloor c'(x);dx $$
Where $lfloor xrfloor$ is the floor function.
Hence, using
$$lfloor xrfloor = x -{x} $$
where ${x}$ is the fractional part of $x$, the partial sum becomes
$$s_{n} = n c_n -sum_{k=1}^{n-1} int_{k}^{k+1} lfloor xrfloor c'(x);dx \
= n c_n -int_{1}^{n} x c'(x);dx + int_{1}^{n} {x} c'(x);dx$$
By partial integration of the first integral the term $n c_{n}$ drops out and we get
$$s_{n} = cos(1) +int_{1}^{n} c(x);dx + int_{1}^{n} {x} c'(x);dx$$
The first integral can be solved explicitly
$$int_{1}^{n} c(x);dx = 2 text{Ci}left(sqrt{n}right)-2 text{Ci}(1)$$
Where $text{Ci}$ is the integral cosine.
The second integral is absolutely convergent and can be very roughly estimated thus (notice that $0le {x} lt 1$)
$$|i_2| = |int_{1}^{n} {x} c'(x),dx| <= int_{1}^{n}| {x} c'(x)|,dx \
<= int_{1}^{n} |{x}| |c'(x)|,dx
<=int_{1}^{n} | c'(x) | ,dx$$
Now
$$| c'(x) | = |frac{cos left(sqrt{x}right)}{x^2}+frac{sin left(sqrt{x}right)}{2 x^{3/2}}| \
leq | frac{sin left(sqrt{x}right)}{2 x^{3/2}}| +|frac{cos left(sqrt{x}right)}{x^2}| leq frac{1}{2 x^{3/2}}+frac{1}{x^2}$$
Hence
$$| i_2| <=int_1^n left(frac{1}{2 x^{3/2}}+frac{1}{x^2}right) , dx = 2-frac{sqrt{n}+1}{n}tag{*}$$
Since $lim_{nto infty } , text{Ci}left(sqrt{n}right)= 0$ the limit of the partial sum is given by
$$s = cos(1) - 2 text{Ci}(1) + lim_{nto infty } ,i_2$$
Observing (*) $s$ remains finite as $ntoinfty$ hence the original sum is convergent. QED.
Remark
A more sophisticated study if the integral $i_2$ might provide better numerical bounds, and even a closed form.
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $d lfloor x rfloor$.
Letting
$$c_{k} = c(k) = frac{cos(sqrt{k})}{k}$$
we attempt to find an integral representation for the partial sum
$$s_n = sum_{k=1}^n c_{k}$$
The formulas for partial summation are
$$sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$
$$A_{k} = sum_{i=1}^k a_{i}$$
Letting $a_{k}=1, b_{k} = c_{k}$ we have $A_{k} = k$.
Now comes the trick which introduces an integral: we have
$$(b_{k}-b_{k+1}) = - int_{k}^{k+1} c'(x);dx$$
and, what's more, the factor $k$ can be incorporated in the integral:
$$k (b_{k}-b_{k+1}) = - k int_{k}^{k+1} c'(x);dx = - int_{k}^{k+1} lfloor xrfloor c'(x);dx $$
Where $lfloor xrfloor$ is the floor function.
Hence, using
$$lfloor xrfloor = x -{x} $$
where ${x}$ is the fractional part of $x$, the partial sum becomes
$$s_{n} = n c_n -sum_{k=1}^{n-1} int_{k}^{k+1} lfloor xrfloor c'(x);dx \
= n c_n -int_{1}^{n} x c'(x);dx + int_{1}^{n} {x} c'(x);dx$$
By partial integration of the first integral the term $n c_{n}$ drops out and we get
$$s_{n} = cos(1) +int_{1}^{n} c(x);dx + int_{1}^{n} {x} c'(x);dx$$
The first integral can be solved explicitly
$$int_{1}^{n} c(x);dx = 2 text{Ci}left(sqrt{n}right)-2 text{Ci}(1)$$
Where $text{Ci}$ is the integral cosine.
The second integral is absolutely convergent and can be very roughly estimated thus (notice that $0le {x} lt 1$)
$$|i_2| = |int_{1}^{n} {x} c'(x),dx| <= int_{1}^{n}| {x} c'(x)|,dx \
<= int_{1}^{n} |{x}| |c'(x)|,dx
<=int_{1}^{n} | c'(x) | ,dx$$
Now
$$| c'(x) | = |frac{cos left(sqrt{x}right)}{x^2}+frac{sin left(sqrt{x}right)}{2 x^{3/2}}| \
leq | frac{sin left(sqrt{x}right)}{2 x^{3/2}}| +|frac{cos left(sqrt{x}right)}{x^2}| leq frac{1}{2 x^{3/2}}+frac{1}{x^2}$$
Hence
$$| i_2| <=int_1^n left(frac{1}{2 x^{3/2}}+frac{1}{x^2}right) , dx = 2-frac{sqrt{n}+1}{n}tag{*}$$
Since $lim_{nto infty } , text{Ci}left(sqrt{n}right)= 0$ the limit of the partial sum is given by
$$s = cos(1) - 2 text{Ci}(1) + lim_{nto infty } ,i_2$$
Observing (*) $s$ remains finite as $ntoinfty$ hence the original sum is convergent. QED.
Remark
A more sophisticated study if the integral $i_2$ might provide better numerical bounds, and even a closed form.
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
edited Oct 11 '18 at 11:04
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
answered Oct 10 '18 at 15:06
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
3,890621
add a comment |
add a comment |
$begingroup$
This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.
Step 1: Transformation to an alternating sum
The partial sum in question is
$$s_n = sum_{k=1}^n c_ktag{1}$$
where
$$c_k = frac{cos(sqrt{k})}{k}tag{2}$$
Now collecting all subsequent summands with the same sign we can write
$$s_n = sum_{m=0}^M (-1)^m f_mtag{3}$$
where
$$f_0 = sum_{k=1}^{lfloor z_{1}rfloor} c_{k}tag{4a}$$
$$f_{mge 1} = (-1)^m sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} c_{k}tag{4b}$$
is the sum of terms between two adjacent roots $k=z_m$ and $k=z_{m+1}$ of $c(k) = 0$, and $M$ is some number depending on $n$ which can in principle be specified but it is normally not necessary as it goes to $infty$ together with $n$
Then, by an appropirate choice of the factor $(-1)^m$, all $f_m$ can be chosen to be positive quantities.
Step 2: Application of Dirichlet's criterion
Now we can apply the Dirichlet criterion to (3). This reads now as follows: $s_n$ is convergent if
(1) $f_m to 0$ for $mtoinfty$
(2) $f_m$ is monotonous
ad (1)
We have for $mge1$
$$|f_{m}| le g(m)$$
where we have dropped the $cos$ and have defined
$$g(m) = sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} frac{1}{k}
=H_{lfloor z_{m+1} rfloor}-H_{lceil z_{m}rceil-1}tag{5} $$
Here $H_n$ ist the harmonic number.
Now the zeroes of $c_{k}$ are given by
$$z_m = left(pi(m-frac{1}{2})right)^2tag{6}$$
Dropping Floor and Ceiling in $g(m)$ defines
$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$
Letting ${lfloor x rfloor} to x-1,{lceil x rceil} to x+1 $ defines another function
$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$
and we have the inequality
$$g_1(m) lt g(m) lt g_0(m)$$
Asymptotically this leads, up to $O(frac{1}{m^2})$, to the inequality
$$frac{2}{m}- frac{1}{pi^2 m^2}< g(m) <frac{2}{m}+frac{1}{pi^2 m^2} $$
This shows that $f_m$ goes to zero.
ad (2)
(to be continued)
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
add a comment |
$begingroup$
This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.
Step 1: Transformation to an alternating sum
The partial sum in question is
$$s_n = sum_{k=1}^n c_ktag{1}$$
where
$$c_k = frac{cos(sqrt{k})}{k}tag{2}$$
Now collecting all subsequent summands with the same sign we can write
$$s_n = sum_{m=0}^M (-1)^m f_mtag{3}$$
where
$$f_0 = sum_{k=1}^{lfloor z_{1}rfloor} c_{k}tag{4a}$$
$$f_{mge 1} = (-1)^m sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} c_{k}tag{4b}$$
is the sum of terms between two adjacent roots $k=z_m$ and $k=z_{m+1}$ of $c(k) = 0$, and $M$ is some number depending on $n$ which can in principle be specified but it is normally not necessary as it goes to $infty$ together with $n$
Then, by an appropirate choice of the factor $(-1)^m$, all $f_m$ can be chosen to be positive quantities.
Step 2: Application of Dirichlet's criterion
Now we can apply the Dirichlet criterion to (3). This reads now as follows: $s_n$ is convergent if
(1) $f_m to 0$ for $mtoinfty$
(2) $f_m$ is monotonous
ad (1)
We have for $mge1$
$$|f_{m}| le g(m)$$
where we have dropped the $cos$ and have defined
$$g(m) = sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} frac{1}{k}
=H_{lfloor z_{m+1} rfloor}-H_{lceil z_{m}rceil-1}tag{5} $$
Here $H_n$ ist the harmonic number.
Now the zeroes of $c_{k}$ are given by
$$z_m = left(pi(m-frac{1}{2})right)^2tag{6}$$
Dropping Floor and Ceiling in $g(m)$ defines
$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$
Letting ${lfloor x rfloor} to x-1,{lceil x rceil} to x+1 $ defines another function
$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$
and we have the inequality
$$g_1(m) lt g(m) lt g_0(m)$$
Asymptotically this leads, up to $O(frac{1}{m^2})$, to the inequality
$$frac{2}{m}- frac{1}{pi^2 m^2}< g(m) <frac{2}{m}+frac{1}{pi^2 m^2} $$
This shows that $f_m$ goes to zero.
ad (2)
(to be continued)
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
add a comment |
$begingroup$
This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.
Step 1: Transformation to an alternating sum
The partial sum in question is
$$s_n = sum_{k=1}^n c_ktag{1}$$
where
$$c_k = frac{cos(sqrt{k})}{k}tag{2}$$
Now collecting all subsequent summands with the same sign we can write
$$s_n = sum_{m=0}^M (-1)^m f_mtag{3}$$
where
$$f_0 = sum_{k=1}^{lfloor z_{1}rfloor} c_{k}tag{4a}$$
$$f_{mge 1} = (-1)^m sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} c_{k}tag{4b}$$
is the sum of terms between two adjacent roots $k=z_m$ and $k=z_{m+1}$ of $c(k) = 0$, and $M$ is some number depending on $n$ which can in principle be specified but it is normally not necessary as it goes to $infty$ together with $n$
Then, by an appropirate choice of the factor $(-1)^m$, all $f_m$ can be chosen to be positive quantities.
Step 2: Application of Dirichlet's criterion
Now we can apply the Dirichlet criterion to (3). This reads now as follows: $s_n$ is convergent if
(1) $f_m to 0$ for $mtoinfty$
(2) $f_m$ is monotonous
ad (1)
We have for $mge1$
$$|f_{m}| le g(m)$$
where we have dropped the $cos$ and have defined
$$g(m) = sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} frac{1}{k}
=H_{lfloor z_{m+1} rfloor}-H_{lceil z_{m}rceil-1}tag{5} $$
Here $H_n$ ist the harmonic number.
Now the zeroes of $c_{k}$ are given by
$$z_m = left(pi(m-frac{1}{2})right)^2tag{6}$$
Dropping Floor and Ceiling in $g(m)$ defines
$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$
Letting ${lfloor x rfloor} to x-1,{lceil x rceil} to x+1 $ defines another function
$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$
and we have the inequality
$$g_1(m) lt g(m) lt g_0(m)$$
Asymptotically this leads, up to $O(frac{1}{m^2})$, to the inequality
$$frac{2}{m}- frac{1}{pi^2 m^2}< g(m) <frac{2}{m}+frac{1}{pi^2 m^2} $$
This shows that $f_m$ goes to zero.
ad (2)
(to be continued)
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
$endgroup$
This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.
Step 1: Transformation to an alternating sum
The partial sum in question is
$$s_n = sum_{k=1}^n c_ktag{1}$$
where
$$c_k = frac{cos(sqrt{k})}{k}tag{2}$$
Now collecting all subsequent summands with the same sign we can write
$$s_n = sum_{m=0}^M (-1)^m f_mtag{3}$$
where
$$f_0 = sum_{k=1}^{lfloor z_{1}rfloor} c_{k}tag{4a}$$
$$f_{mge 1} = (-1)^m sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} c_{k}tag{4b}$$
is the sum of terms between two adjacent roots $k=z_m$ and $k=z_{m+1}$ of $c(k) = 0$, and $M$ is some number depending on $n$ which can in principle be specified but it is normally not necessary as it goes to $infty$ together with $n$
Then, by an appropirate choice of the factor $(-1)^m$, all $f_m$ can be chosen to be positive quantities.
Step 2: Application of Dirichlet's criterion
Now we can apply the Dirichlet criterion to (3). This reads now as follows: $s_n$ is convergent if
(1) $f_m to 0$ for $mtoinfty$
(2) $f_m$ is monotonous
ad (1)
We have for $mge1$
$$|f_{m}| le g(m)$$
where we have dropped the $cos$ and have defined
$$g(m) = sum_{k={lceil z_{m}rceil}}^{lfloor z_{m+1} rfloor} frac{1}{k}
=H_{lfloor z_{m+1} rfloor}-H_{lceil z_{m}rceil-1}tag{5} $$
Here $H_n$ ist the harmonic number.
Now the zeroes of $c_{k}$ are given by
$$z_m = left(pi(m-frac{1}{2})right)^2tag{6}$$
Dropping Floor and Ceiling in $g(m)$ defines
$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$
Letting ${lfloor x rfloor} to x-1,{lceil x rceil} to x+1 $ defines another function
$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$
and we have the inequality
$$g_1(m) lt g(m) lt g_0(m)$$
Asymptotically this leads, up to $O(frac{1}{m^2})$, to the inequality
$$frac{2}{m}- frac{1}{pi^2 m^2}< g(m) <frac{2}{m}+frac{1}{pi^2 m^2} $$
This shows that $f_m$ goes to zero.
ad (2)
(to be continued)
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
edited Oct 25 '18 at 8:56
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
answered Oct 24 '18 at 15:27
Dr. Wolfgang HintzeDr. Wolfgang Hintze
3,890621
3,890621
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$begingroup$
If you approximate the sum by an integral, then you have $$sum_{n=1}^m,frac{cos(sqrt{n})}{n}approx int_1^m,frac{cos(sqrt{x})}{x},text{d}x=2,text{Ci}(sqrt{m})-2,text{Ci}(1),,$$ where $text{Ci}$ is the cosine integral $$text{Ci}(t)=-int_t^infty,frac{cos(s)}{s},text{d}s,.$$ However, I do not know how accurate the approximation is. But I am convinced that the sum converges since $$lim_{mtoinfty},text{Ci}(sqrt{m})=0,.$$
$endgroup$
– Batominovski
Oct 9 '18 at 11:27
$begingroup$
Plus, after trying to sum up to $3000$ terms, it seems to me that the sum fluctuates around $-0.32$. This looks very promising that the sum does converge to a limit around $-0.32$.
$endgroup$
– Batominovski
Oct 9 '18 at 11:32