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$sum_{n=1}^infty a_n cos nx$ unbounded near $0$ if $sum a_n$ diverges?
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If $a_n$ is a decreasing positive sequence and tends to $0$, and given$$sum_{n=1}^infty a_n=+infty$$
can we prove that $$lim_{xrightarrow 0}sum_{n=1}^{infty} a_n cos nx =+infty$$
or at least prove the series above is unbounded for $x$ in a neighborhood of $0$?
For the power series $$sum_{n=1}^infty a_n left(1-xright)^n $$ the conclusion holds, because $left(1-xright)^n$ is positive.
I wonder if trigonometric series can have the similar conclusion, so I tested several $a_n$, plotted the graph, and discovered that it is probably true.
However, since $cos nx$ is not identically positive, it is hard to give a rigorous proof, and I cannot give a counter-example either.
Anyone has some ideas?
real-analysis sequences-and-series complex-analysis analysis trigonometric-series
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
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add a comment |
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If $a_n$ is a decreasing positive sequence and tends to $0$, and given$$sum_{n=1}^infty a_n=+infty$$
can we prove that $$lim_{xrightarrow 0}sum_{n=1}^{infty} a_n cos nx =+infty$$
or at least prove the series above is unbounded for $x$ in a neighborhood of $0$?
For the power series $$sum_{n=1}^infty a_n left(1-xright)^n $$ the conclusion holds, because $left(1-xright)^n$ is positive.
I wonder if trigonometric series can have the similar conclusion, so I tested several $a_n$, plotted the graph, and discovered that it is probably true.
However, since $cos nx$ is not identically positive, it is hard to give a rigorous proof, and I cannot give a counter-example either.
Anyone has some ideas?
real-analysis sequences-and-series complex-analysis analysis trigonometric-series
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
$endgroup$
add a comment |
$begingroup$
If $a_n$ is a decreasing positive sequence and tends to $0$, and given$$sum_{n=1}^infty a_n=+infty$$
can we prove that $$lim_{xrightarrow 0}sum_{n=1}^{infty} a_n cos nx =+infty$$
or at least prove the series above is unbounded for $x$ in a neighborhood of $0$?
For the power series $$sum_{n=1}^infty a_n left(1-xright)^n $$ the conclusion holds, because $left(1-xright)^n$ is positive.
I wonder if trigonometric series can have the similar conclusion, so I tested several $a_n$, plotted the graph, and discovered that it is probably true.
However, since $cos nx$ is not identically positive, it is hard to give a rigorous proof, and I cannot give a counter-example either.
Anyone has some ideas?
real-analysis sequences-and-series complex-analysis analysis trigonometric-series
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
$endgroup$
If $a_n$ is a decreasing positive sequence and tends to $0$, and given$$sum_{n=1}^infty a_n=+infty$$
can we prove that $$lim_{xrightarrow 0}sum_{n=1}^{infty} a_n cos nx =+infty$$
or at least prove the series above is unbounded for $x$ in a neighborhood of $0$?
For the power series $$sum_{n=1}^infty a_n left(1-xright)^n $$ the conclusion holds, because $left(1-xright)^n$ is positive.
I wonder if trigonometric series can have the similar conclusion, so I tested several $a_n$, plotted the graph, and discovered that it is probably true.
However, since $cos nx$ is not identically positive, it is hard to give a rigorous proof, and I cannot give a counter-example either.
Anyone has some ideas?
real-analysis sequences-and-series complex-analysis analysis trigonometric-series
real-analysis sequences-and-series complex-analysis analysis trigonometric-series
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
asked Jan 26 at 2:09
AntimoniusAntimonius
398110
398110
add a comment |
add a comment |
2 Answers
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Now I can answer half of the question.
The conclusion is as below:
Theorem 1. For monotone decreasing sequence $a_n$ satisfying $a_nrightarrow 0$ and $sum a_n$ diverges, the sum $$sum_{n=1}^infty a_ncos nx $$ must be UNBOUNDED in a neighborhood of $0$.
The proof goes like this:
First we have $a_nge 0$. By Dirichlet’s method, the series $$Sleft(xright)=sum_{n=1}^infty a_ncos nx $$
converges uniformly on $left(delta,piright]$ for any $delta>0$, therefore is continuous on it.
If $S$ is not integrable on $left[0,piright]$, it must be unbounded on $left(0,deltaright)$ for some $delta>0$.
If $S$ is integrable on $left[0,piright]$, then the Fourier series of $S$ is $Sleft(xright)$ itself (See Hardy Fourier Series §3.10 Thm.46).
By the properties of Fejér kernel, we can write the n-th Cesàro partial sum of $S$ at $x=0$ as $$sigma_n=frac{1}{pi}int_0^pi Sleft(tright)F_nleft(tright)dt$$
where $F_nge 0$ is the Fejér kernel.
Suppose $left|Sleft(xright)right|le M$, then $$sigma_nle frac{M}{pi}int_0^pi F_nleft(tright)dt=M$$
As the series $sum a_n$ is positive and divergent, the n-th Cesàro partial sum $sigma_nrightarrow+infty$ as $nrightarrow infty$(by Stolz’s theorem), which leads to contradiction.
In conclusion, $S$ must be unbounded near $0$. $blacksquare$
But it remains to prove or disprove $Sleft(xright)rightarrow+infty$ as $xrightarrow0$. It still needs tougher work.
There are also two little theorem about divergence to $+infty$ of $Sleft(xright)$:
Theorem 2. If $a_n$ is a CONVEX sequence(i.e.$a_n+a_{n+2}ge 2a_{n+1},forall n$) tends to $0$, and $sum a_n$ diverges, then $Sleft(xright)rightarrow +infty$ when $xrightarrow 0$.
Theorem 3. If $a_n$ is defined as Thm.1, and $S$ is NOT integrable on $left[0,piright]$, $S$ must have infinitely many zeroes near $0$, hence does not tend to $+infty$.
Proof(Thm.2):
Using summation by parts twice and we can get $$Sleft(xright)=sum_{n=1}^infty left(a_n+a_{n+2}-2a_{n+1}right)frac{1-cosleft(n+1right)x}{4sin^2 frac{x}{2}}$$
which is a positive-term series.
There exists a constant $C>0$, for $0<x<frac{1}{n}$, $$frac{1-cos nx}{4sin^2 frac{x}{2}}ge Cn^2$$
Therefore $$Sleft(xright)ge Csum_{n=1}^{left[frac{1}{x}right]-1} left(n+1right)^2 left(a_n+a_{n+2}-2a_{n+1}right)$$
Then we need to prove that RHS is divergent.
We need a lemma: If $b_n$ is decreasing and tends to $0$, $sum n^k b_n$ diverges, then $sum n^{k+1}left(b_n-b_{n+1}right)$ also diverges.
Since $$Delta n^{k+1}=n^{k+1}-left(n-1right)^{k+1}sim left(k+1right) n^k$$ we know that $sum Delta n^{k+1},b_n$ diverges.
For $M>0$, There exists $N_0$ s.t. $$sum_{n=1}^{N_0}Delta n^{k+1},b_n>M+1$$
As $b_nrightarrow 0$, there exists $N_1$, for all $N>N_1$ we have $N_0^{k+1}b_N<1$.
Then for $N>maxleft{N_0,N_1right}$,$$begin{aligned}sum_{n=1}^N n^{k+1}left(b_n-b_{n+1}right)&=sum_{n=1}^N Delta n^{k+1}left(b_n-b_{N+1}right)\&gesum_{n=1}^{N_0}Delta n^{k+1} left(b_n-b_{N+1}right)\&=sum_{n=1}^{N_0} Delta n^{k+1},b_n -N_0^{k+1} b_{N+1}\&>M+1-1=Mend{aligned}$$
Hence by the lemma we can get $$sum a_n =+infty Rightarrow sum n left(a_n-a_{n+1}right) =+infty Rightarrow sum n^2 left(a_n+a_{n+2}-2a_{n+1}right)=+infty$$
which finally proves the theorem. $blacksquare $
The proof of Thm.3 needs two lemmas as follow.
Lemma 1. If $b_n$ is decreasing and $nb_nrightarrow 0$, then $sum_{n=1}^infty b_n sin nx $ converges uniformly on $mathbb{R}$.
Lemma 2. $S$ has an antiderivative $sum _{n=1}^infty frac{a_n}{n}sin nx$ and $$lim_{deltarightarrow 0^+}int _delta ^pi 2Sleft(xright) cos mx dx=pi a_m$$
The proof of Lemma.1 is a little complex but not difficult, and is taken from an exercise book in mathematical analysis. We will just assume it right.
Proof of Lemma.2:
Notice that the series of $S$ converges uniformly in $left[delta,piright]$, hence we can integrate by terms $$begin{aligned}int_delta^pi 2Sleft(xright)cos mx,dx&=sum_{n=1}^infty int_delta^pi 2a_n cos nx cos mx,dx\&=sum_{n=1}^infty frac{a_n}{n+m}sinleft(n+mright)delta +sum_{nne m}frac{a_n}{n-m}sin left(n-mright)delta +a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)\&overset{def}{=}Theta_1left(deltaright)+Theta_2left(deltaright)+a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)end{aligned}$$
Clearly by lemma 1, $Theta_1$ and $Theta _2$ both converge uniformly on $mathbb{R}$, hence continuous at 0.
So we have $$lim_{deltarightarrow 0^+}int_delta^pi 2Sleft(xright)cos mx,dx=pi a_m$$
Letting $m=0$ we can get that $$int_x^pi Sleft(tright) dt=sum_{n=1}^infty frac{a_n}{n}sin nx$$
By the continuity of $S$ we finish the proof. $blacksquare$
Back to the proof of Thm.3
Note that $S$ is not integrable, that means $$int_0^pi left|Sleft(tright)right| dt= +infty$$
But the improper integral of $S$ converges to $0$ as discussed above.
That means S must have infinitely many positive parts and negative parts near $0$, by continuity there must be infinitely many zeroes. $blacksquare$
So if we are going to find $S$ such that $lim_{xrightarrow 0}Sleft(xright)ne +infty$, we can just find an unintegrable $S$, and the coefficient $a_n$ must not be convex.
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
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Would $S’(x)$ give some useful information?
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– Szeto
Jan 28 at 23:20
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@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
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– Antimonius
Jan 29 at 3:48
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All I can think of at this moment is the following very restricted partial result:
Claim. If $(a_n)$ is non-negative, non-increasing and $n a_n leq C$ for some constant $C > 0$, then
$$ lim_{ntoinfty} sum_{n=0}^{infty} a_n cos(n x) = sum_{n=0}^{infty} a_n, $$
regradless of the convergence of $sum_{n=0}^{infty} a_n$.
Since the claim is obvious if $sum_{n=0}^{infty} a_n < infty$, we may focus on the case $sum_{n=0}^{infty} a_n = infty$. To this end, define an auxiliary sequence $(b_n)$ by $a_n - a_{n+1} = b_n$. Then
$b_n geq 0$ and $a_m = a_n + sum_{k=m}^{n-1} b_k$ for $0 leq m < n$,
$a_n = sum_{k=n}^{infty} b_k$, and
$sum_{k=0}^{infty} (k+1)b_k = sum_{n=0}^{infty} a_n = infty$. (This follows from Tonelli's theorem.)
Moreover, we can apply summation by parts
begin{align*}
sum_{n=0}^{N} a_n cos(nx)
&= a_N sum_{n=0}^{N} cos(nx) + sum_{n=0}^{N-1} left( sum_{k=n}^{N-1} b_k right) cos(nx) \
&= a_N D_N(x) + sum_{k=0}^{N-1} b_k D_k(x),
end{align*}
where $D_k(x) = sum_{n=0}^{k} cos(nx)$. Now if $x in (0, pi]$, then $D_n(x)$ is bounded in $n$. So, as $Ntoinfty$, the above converges to
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
= sum_{k=0}^{infty} b_k D_k(x).
end{align*}
Now notice that $D_k(x) = cos(kx/2)sin((k+1)x/2)/sin(x/2)$. So, if $x in (0, 1)$ and $k leq 1/x$, then
$$ D_k(x) geq frac{cos(1/2) cdot frac{2}{pi} (k+1)x/2}{x/2} = c(k+1) $$
for $c = frac{2}{pi}cos(1/2) > 0$. So
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
&geq sum_{0 leq k leq 1/x} b_k D_k(x) - sum_{k > 1/x} frac{b_k}{sin(x/2)} \
&geq c sum_{0 leq k leq 1/x} (k+1)b_k - frac{a_{lceil 1/x rceil}}{sin(x/2)}.
end{align*}
Since $a_{lceil 1/x rceil} = mathcal{O}(x)$ as $x to 0^+$, we have $frac{a_{lceil 1/x rceil}}{sin(x/2)} = mathcal{O}(1)$, and so, letting $xto 0^+$ proves the desired claim.
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
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Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
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– Antimonius
Feb 1 at 3:18
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2 Answers
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2 Answers
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$begingroup$
Now I can answer half of the question.
The conclusion is as below:
Theorem 1. For monotone decreasing sequence $a_n$ satisfying $a_nrightarrow 0$ and $sum a_n$ diverges, the sum $$sum_{n=1}^infty a_ncos nx $$ must be UNBOUNDED in a neighborhood of $0$.
The proof goes like this:
First we have $a_nge 0$. By Dirichlet’s method, the series $$Sleft(xright)=sum_{n=1}^infty a_ncos nx $$
converges uniformly on $left(delta,piright]$ for any $delta>0$, therefore is continuous on it.
If $S$ is not integrable on $left[0,piright]$, it must be unbounded on $left(0,deltaright)$ for some $delta>0$.
If $S$ is integrable on $left[0,piright]$, then the Fourier series of $S$ is $Sleft(xright)$ itself (See Hardy Fourier Series §3.10 Thm.46).
By the properties of Fejér kernel, we can write the n-th Cesàro partial sum of $S$ at $x=0$ as $$sigma_n=frac{1}{pi}int_0^pi Sleft(tright)F_nleft(tright)dt$$
where $F_nge 0$ is the Fejér kernel.
Suppose $left|Sleft(xright)right|le M$, then $$sigma_nle frac{M}{pi}int_0^pi F_nleft(tright)dt=M$$
As the series $sum a_n$ is positive and divergent, the n-th Cesàro partial sum $sigma_nrightarrow+infty$ as $nrightarrow infty$(by Stolz’s theorem), which leads to contradiction.
In conclusion, $S$ must be unbounded near $0$. $blacksquare$
But it remains to prove or disprove $Sleft(xright)rightarrow+infty$ as $xrightarrow0$. It still needs tougher work.
There are also two little theorem about divergence to $+infty$ of $Sleft(xright)$:
Theorem 2. If $a_n$ is a CONVEX sequence(i.e.$a_n+a_{n+2}ge 2a_{n+1},forall n$) tends to $0$, and $sum a_n$ diverges, then $Sleft(xright)rightarrow +infty$ when $xrightarrow 0$.
Theorem 3. If $a_n$ is defined as Thm.1, and $S$ is NOT integrable on $left[0,piright]$, $S$ must have infinitely many zeroes near $0$, hence does not tend to $+infty$.
Proof(Thm.2):
Using summation by parts twice and we can get $$Sleft(xright)=sum_{n=1}^infty left(a_n+a_{n+2}-2a_{n+1}right)frac{1-cosleft(n+1right)x}{4sin^2 frac{x}{2}}$$
which is a positive-term series.
There exists a constant $C>0$, for $0<x<frac{1}{n}$, $$frac{1-cos nx}{4sin^2 frac{x}{2}}ge Cn^2$$
Therefore $$Sleft(xright)ge Csum_{n=1}^{left[frac{1}{x}right]-1} left(n+1right)^2 left(a_n+a_{n+2}-2a_{n+1}right)$$
Then we need to prove that RHS is divergent.
We need a lemma: If $b_n$ is decreasing and tends to $0$, $sum n^k b_n$ diverges, then $sum n^{k+1}left(b_n-b_{n+1}right)$ also diverges.
Since $$Delta n^{k+1}=n^{k+1}-left(n-1right)^{k+1}sim left(k+1right) n^k$$ we know that $sum Delta n^{k+1},b_n$ diverges.
For $M>0$, There exists $N_0$ s.t. $$sum_{n=1}^{N_0}Delta n^{k+1},b_n>M+1$$
As $b_nrightarrow 0$, there exists $N_1$, for all $N>N_1$ we have $N_0^{k+1}b_N<1$.
Then for $N>maxleft{N_0,N_1right}$,$$begin{aligned}sum_{n=1}^N n^{k+1}left(b_n-b_{n+1}right)&=sum_{n=1}^N Delta n^{k+1}left(b_n-b_{N+1}right)\&gesum_{n=1}^{N_0}Delta n^{k+1} left(b_n-b_{N+1}right)\&=sum_{n=1}^{N_0} Delta n^{k+1},b_n -N_0^{k+1} b_{N+1}\&>M+1-1=Mend{aligned}$$
Hence by the lemma we can get $$sum a_n =+infty Rightarrow sum n left(a_n-a_{n+1}right) =+infty Rightarrow sum n^2 left(a_n+a_{n+2}-2a_{n+1}right)=+infty$$
which finally proves the theorem. $blacksquare $
The proof of Thm.3 needs two lemmas as follow.
Lemma 1. If $b_n$ is decreasing and $nb_nrightarrow 0$, then $sum_{n=1}^infty b_n sin nx $ converges uniformly on $mathbb{R}$.
Lemma 2. $S$ has an antiderivative $sum _{n=1}^infty frac{a_n}{n}sin nx$ and $$lim_{deltarightarrow 0^+}int _delta ^pi 2Sleft(xright) cos mx dx=pi a_m$$
The proof of Lemma.1 is a little complex but not difficult, and is taken from an exercise book in mathematical analysis. We will just assume it right.
Proof of Lemma.2:
Notice that the series of $S$ converges uniformly in $left[delta,piright]$, hence we can integrate by terms $$begin{aligned}int_delta^pi 2Sleft(xright)cos mx,dx&=sum_{n=1}^infty int_delta^pi 2a_n cos nx cos mx,dx\&=sum_{n=1}^infty frac{a_n}{n+m}sinleft(n+mright)delta +sum_{nne m}frac{a_n}{n-m}sin left(n-mright)delta +a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)\&overset{def}{=}Theta_1left(deltaright)+Theta_2left(deltaright)+a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)end{aligned}$$
Clearly by lemma 1, $Theta_1$ and $Theta _2$ both converge uniformly on $mathbb{R}$, hence continuous at 0.
So we have $$lim_{deltarightarrow 0^+}int_delta^pi 2Sleft(xright)cos mx,dx=pi a_m$$
Letting $m=0$ we can get that $$int_x^pi Sleft(tright) dt=sum_{n=1}^infty frac{a_n}{n}sin nx$$
By the continuity of $S$ we finish the proof. $blacksquare$
Back to the proof of Thm.3
Note that $S$ is not integrable, that means $$int_0^pi left|Sleft(tright)right| dt= +infty$$
But the improper integral of $S$ converges to $0$ as discussed above.
That means S must have infinitely many positive parts and negative parts near $0$, by continuity there must be infinitely many zeroes. $blacksquare$
So if we are going to find $S$ such that $lim_{xrightarrow 0}Sleft(xright)ne +infty$, we can just find an unintegrable $S$, and the coefficient $a_n$ must not be convex.
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
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$begingroup$
Would $S’(x)$ give some useful information?
$endgroup$
– Szeto
Jan 28 at 23:20
$begingroup$
@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
$endgroup$
– Antimonius
Jan 29 at 3:48
add a comment |
$begingroup$
Now I can answer half of the question.
The conclusion is as below:
Theorem 1. For monotone decreasing sequence $a_n$ satisfying $a_nrightarrow 0$ and $sum a_n$ diverges, the sum $$sum_{n=1}^infty a_ncos nx $$ must be UNBOUNDED in a neighborhood of $0$.
The proof goes like this:
First we have $a_nge 0$. By Dirichlet’s method, the series $$Sleft(xright)=sum_{n=1}^infty a_ncos nx $$
converges uniformly on $left(delta,piright]$ for any $delta>0$, therefore is continuous on it.
If $S$ is not integrable on $left[0,piright]$, it must be unbounded on $left(0,deltaright)$ for some $delta>0$.
If $S$ is integrable on $left[0,piright]$, then the Fourier series of $S$ is $Sleft(xright)$ itself (See Hardy Fourier Series §3.10 Thm.46).
By the properties of Fejér kernel, we can write the n-th Cesàro partial sum of $S$ at $x=0$ as $$sigma_n=frac{1}{pi}int_0^pi Sleft(tright)F_nleft(tright)dt$$
where $F_nge 0$ is the Fejér kernel.
Suppose $left|Sleft(xright)right|le M$, then $$sigma_nle frac{M}{pi}int_0^pi F_nleft(tright)dt=M$$
As the series $sum a_n$ is positive and divergent, the n-th Cesàro partial sum $sigma_nrightarrow+infty$ as $nrightarrow infty$(by Stolz’s theorem), which leads to contradiction.
In conclusion, $S$ must be unbounded near $0$. $blacksquare$
But it remains to prove or disprove $Sleft(xright)rightarrow+infty$ as $xrightarrow0$. It still needs tougher work.
There are also two little theorem about divergence to $+infty$ of $Sleft(xright)$:
Theorem 2. If $a_n$ is a CONVEX sequence(i.e.$a_n+a_{n+2}ge 2a_{n+1},forall n$) tends to $0$, and $sum a_n$ diverges, then $Sleft(xright)rightarrow +infty$ when $xrightarrow 0$.
Theorem 3. If $a_n$ is defined as Thm.1, and $S$ is NOT integrable on $left[0,piright]$, $S$ must have infinitely many zeroes near $0$, hence does not tend to $+infty$.
Proof(Thm.2):
Using summation by parts twice and we can get $$Sleft(xright)=sum_{n=1}^infty left(a_n+a_{n+2}-2a_{n+1}right)frac{1-cosleft(n+1right)x}{4sin^2 frac{x}{2}}$$
which is a positive-term series.
There exists a constant $C>0$, for $0<x<frac{1}{n}$, $$frac{1-cos nx}{4sin^2 frac{x}{2}}ge Cn^2$$
Therefore $$Sleft(xright)ge Csum_{n=1}^{left[frac{1}{x}right]-1} left(n+1right)^2 left(a_n+a_{n+2}-2a_{n+1}right)$$
Then we need to prove that RHS is divergent.
We need a lemma: If $b_n$ is decreasing and tends to $0$, $sum n^k b_n$ diverges, then $sum n^{k+1}left(b_n-b_{n+1}right)$ also diverges.
Since $$Delta n^{k+1}=n^{k+1}-left(n-1right)^{k+1}sim left(k+1right) n^k$$ we know that $sum Delta n^{k+1},b_n$ diverges.
For $M>0$, There exists $N_0$ s.t. $$sum_{n=1}^{N_0}Delta n^{k+1},b_n>M+1$$
As $b_nrightarrow 0$, there exists $N_1$, for all $N>N_1$ we have $N_0^{k+1}b_N<1$.
Then for $N>maxleft{N_0,N_1right}$,$$begin{aligned}sum_{n=1}^N n^{k+1}left(b_n-b_{n+1}right)&=sum_{n=1}^N Delta n^{k+1}left(b_n-b_{N+1}right)\&gesum_{n=1}^{N_0}Delta n^{k+1} left(b_n-b_{N+1}right)\&=sum_{n=1}^{N_0} Delta n^{k+1},b_n -N_0^{k+1} b_{N+1}\&>M+1-1=Mend{aligned}$$
Hence by the lemma we can get $$sum a_n =+infty Rightarrow sum n left(a_n-a_{n+1}right) =+infty Rightarrow sum n^2 left(a_n+a_{n+2}-2a_{n+1}right)=+infty$$
which finally proves the theorem. $blacksquare $
The proof of Thm.3 needs two lemmas as follow.
Lemma 1. If $b_n$ is decreasing and $nb_nrightarrow 0$, then $sum_{n=1}^infty b_n sin nx $ converges uniformly on $mathbb{R}$.
Lemma 2. $S$ has an antiderivative $sum _{n=1}^infty frac{a_n}{n}sin nx$ and $$lim_{deltarightarrow 0^+}int _delta ^pi 2Sleft(xright) cos mx dx=pi a_m$$
The proof of Lemma.1 is a little complex but not difficult, and is taken from an exercise book in mathematical analysis. We will just assume it right.
Proof of Lemma.2:
Notice that the series of $S$ converges uniformly in $left[delta,piright]$, hence we can integrate by terms $$begin{aligned}int_delta^pi 2Sleft(xright)cos mx,dx&=sum_{n=1}^infty int_delta^pi 2a_n cos nx cos mx,dx\&=sum_{n=1}^infty frac{a_n}{n+m}sinleft(n+mright)delta +sum_{nne m}frac{a_n}{n-m}sin left(n-mright)delta +a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)\&overset{def}{=}Theta_1left(deltaright)+Theta_2left(deltaright)+a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)end{aligned}$$
Clearly by lemma 1, $Theta_1$ and $Theta _2$ both converge uniformly on $mathbb{R}$, hence continuous at 0.
So we have $$lim_{deltarightarrow 0^+}int_delta^pi 2Sleft(xright)cos mx,dx=pi a_m$$
Letting $m=0$ we can get that $$int_x^pi Sleft(tright) dt=sum_{n=1}^infty frac{a_n}{n}sin nx$$
By the continuity of $S$ we finish the proof. $blacksquare$
Back to the proof of Thm.3
Note that $S$ is not integrable, that means $$int_0^pi left|Sleft(tright)right| dt= +infty$$
But the improper integral of $S$ converges to $0$ as discussed above.
That means S must have infinitely many positive parts and negative parts near $0$, by continuity there must be infinitely many zeroes. $blacksquare$
So if we are going to find $S$ such that $lim_{xrightarrow 0}Sleft(xright)ne +infty$, we can just find an unintegrable $S$, and the coefficient $a_n$ must not be convex.
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
$endgroup$
$begingroup$
Would $S’(x)$ give some useful information?
$endgroup$
– Szeto
Jan 28 at 23:20
$begingroup$
@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
$endgroup$
– Antimonius
Jan 29 at 3:48
add a comment |
$begingroup$
Now I can answer half of the question.
The conclusion is as below:
Theorem 1. For monotone decreasing sequence $a_n$ satisfying $a_nrightarrow 0$ and $sum a_n$ diverges, the sum $$sum_{n=1}^infty a_ncos nx $$ must be UNBOUNDED in a neighborhood of $0$.
The proof goes like this:
First we have $a_nge 0$. By Dirichlet’s method, the series $$Sleft(xright)=sum_{n=1}^infty a_ncos nx $$
converges uniformly on $left(delta,piright]$ for any $delta>0$, therefore is continuous on it.
If $S$ is not integrable on $left[0,piright]$, it must be unbounded on $left(0,deltaright)$ for some $delta>0$.
If $S$ is integrable on $left[0,piright]$, then the Fourier series of $S$ is $Sleft(xright)$ itself (See Hardy Fourier Series §3.10 Thm.46).
By the properties of Fejér kernel, we can write the n-th Cesàro partial sum of $S$ at $x=0$ as $$sigma_n=frac{1}{pi}int_0^pi Sleft(tright)F_nleft(tright)dt$$
where $F_nge 0$ is the Fejér kernel.
Suppose $left|Sleft(xright)right|le M$, then $$sigma_nle frac{M}{pi}int_0^pi F_nleft(tright)dt=M$$
As the series $sum a_n$ is positive and divergent, the n-th Cesàro partial sum $sigma_nrightarrow+infty$ as $nrightarrow infty$(by Stolz’s theorem), which leads to contradiction.
In conclusion, $S$ must be unbounded near $0$. $blacksquare$
But it remains to prove or disprove $Sleft(xright)rightarrow+infty$ as $xrightarrow0$. It still needs tougher work.
There are also two little theorem about divergence to $+infty$ of $Sleft(xright)$:
Theorem 2. If $a_n$ is a CONVEX sequence(i.e.$a_n+a_{n+2}ge 2a_{n+1},forall n$) tends to $0$, and $sum a_n$ diverges, then $Sleft(xright)rightarrow +infty$ when $xrightarrow 0$.
Theorem 3. If $a_n$ is defined as Thm.1, and $S$ is NOT integrable on $left[0,piright]$, $S$ must have infinitely many zeroes near $0$, hence does not tend to $+infty$.
Proof(Thm.2):
Using summation by parts twice and we can get $$Sleft(xright)=sum_{n=1}^infty left(a_n+a_{n+2}-2a_{n+1}right)frac{1-cosleft(n+1right)x}{4sin^2 frac{x}{2}}$$
which is a positive-term series.
There exists a constant $C>0$, for $0<x<frac{1}{n}$, $$frac{1-cos nx}{4sin^2 frac{x}{2}}ge Cn^2$$
Therefore $$Sleft(xright)ge Csum_{n=1}^{left[frac{1}{x}right]-1} left(n+1right)^2 left(a_n+a_{n+2}-2a_{n+1}right)$$
Then we need to prove that RHS is divergent.
We need a lemma: If $b_n$ is decreasing and tends to $0$, $sum n^k b_n$ diverges, then $sum n^{k+1}left(b_n-b_{n+1}right)$ also diverges.
Since $$Delta n^{k+1}=n^{k+1}-left(n-1right)^{k+1}sim left(k+1right) n^k$$ we know that $sum Delta n^{k+1},b_n$ diverges.
For $M>0$, There exists $N_0$ s.t. $$sum_{n=1}^{N_0}Delta n^{k+1},b_n>M+1$$
As $b_nrightarrow 0$, there exists $N_1$, for all $N>N_1$ we have $N_0^{k+1}b_N<1$.
Then for $N>maxleft{N_0,N_1right}$,$$begin{aligned}sum_{n=1}^N n^{k+1}left(b_n-b_{n+1}right)&=sum_{n=1}^N Delta n^{k+1}left(b_n-b_{N+1}right)\&gesum_{n=1}^{N_0}Delta n^{k+1} left(b_n-b_{N+1}right)\&=sum_{n=1}^{N_0} Delta n^{k+1},b_n -N_0^{k+1} b_{N+1}\&>M+1-1=Mend{aligned}$$
Hence by the lemma we can get $$sum a_n =+infty Rightarrow sum n left(a_n-a_{n+1}right) =+infty Rightarrow sum n^2 left(a_n+a_{n+2}-2a_{n+1}right)=+infty$$
which finally proves the theorem. $blacksquare $
The proof of Thm.3 needs two lemmas as follow.
Lemma 1. If $b_n$ is decreasing and $nb_nrightarrow 0$, then $sum_{n=1}^infty b_n sin nx $ converges uniformly on $mathbb{R}$.
Lemma 2. $S$ has an antiderivative $sum _{n=1}^infty frac{a_n}{n}sin nx$ and $$lim_{deltarightarrow 0^+}int _delta ^pi 2Sleft(xright) cos mx dx=pi a_m$$
The proof of Lemma.1 is a little complex but not difficult, and is taken from an exercise book in mathematical analysis. We will just assume it right.
Proof of Lemma.2:
Notice that the series of $S$ converges uniformly in $left[delta,piright]$, hence we can integrate by terms $$begin{aligned}int_delta^pi 2Sleft(xright)cos mx,dx&=sum_{n=1}^infty int_delta^pi 2a_n cos nx cos mx,dx\&=sum_{n=1}^infty frac{a_n}{n+m}sinleft(n+mright)delta +sum_{nne m}frac{a_n}{n-m}sin left(n-mright)delta +a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)\&overset{def}{=}Theta_1left(deltaright)+Theta_2left(deltaright)+a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)end{aligned}$$
Clearly by lemma 1, $Theta_1$ and $Theta _2$ both converge uniformly on $mathbb{R}$, hence continuous at 0.
So we have $$lim_{deltarightarrow 0^+}int_delta^pi 2Sleft(xright)cos mx,dx=pi a_m$$
Letting $m=0$ we can get that $$int_x^pi Sleft(tright) dt=sum_{n=1}^infty frac{a_n}{n}sin nx$$
By the continuity of $S$ we finish the proof. $blacksquare$
Back to the proof of Thm.3
Note that $S$ is not integrable, that means $$int_0^pi left|Sleft(tright)right| dt= +infty$$
But the improper integral of $S$ converges to $0$ as discussed above.
That means S must have infinitely many positive parts and negative parts near $0$, by continuity there must be infinitely many zeroes. $blacksquare$
So if we are going to find $S$ such that $lim_{xrightarrow 0}Sleft(xright)ne +infty$, we can just find an unintegrable $S$, and the coefficient $a_n$ must not be convex.
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
$endgroup$
Now I can answer half of the question.
The conclusion is as below:
Theorem 1. For monotone decreasing sequence $a_n$ satisfying $a_nrightarrow 0$ and $sum a_n$ diverges, the sum $$sum_{n=1}^infty a_ncos nx $$ must be UNBOUNDED in a neighborhood of $0$.
The proof goes like this:
First we have $a_nge 0$. By Dirichlet’s method, the series $$Sleft(xright)=sum_{n=1}^infty a_ncos nx $$
converges uniformly on $left(delta,piright]$ for any $delta>0$, therefore is continuous on it.
If $S$ is not integrable on $left[0,piright]$, it must be unbounded on $left(0,deltaright)$ for some $delta>0$.
If $S$ is integrable on $left[0,piright]$, then the Fourier series of $S$ is $Sleft(xright)$ itself (See Hardy Fourier Series §3.10 Thm.46).
By the properties of Fejér kernel, we can write the n-th Cesàro partial sum of $S$ at $x=0$ as $$sigma_n=frac{1}{pi}int_0^pi Sleft(tright)F_nleft(tright)dt$$
where $F_nge 0$ is the Fejér kernel.
Suppose $left|Sleft(xright)right|le M$, then $$sigma_nle frac{M}{pi}int_0^pi F_nleft(tright)dt=M$$
As the series $sum a_n$ is positive and divergent, the n-th Cesàro partial sum $sigma_nrightarrow+infty$ as $nrightarrow infty$(by Stolz’s theorem), which leads to contradiction.
In conclusion, $S$ must be unbounded near $0$. $blacksquare$
But it remains to prove or disprove $Sleft(xright)rightarrow+infty$ as $xrightarrow0$. It still needs tougher work.
There are also two little theorem about divergence to $+infty$ of $Sleft(xright)$:
Theorem 2. If $a_n$ is a CONVEX sequence(i.e.$a_n+a_{n+2}ge 2a_{n+1},forall n$) tends to $0$, and $sum a_n$ diverges, then $Sleft(xright)rightarrow +infty$ when $xrightarrow 0$.
Theorem 3. If $a_n$ is defined as Thm.1, and $S$ is NOT integrable on $left[0,piright]$, $S$ must have infinitely many zeroes near $0$, hence does not tend to $+infty$.
Proof(Thm.2):
Using summation by parts twice and we can get $$Sleft(xright)=sum_{n=1}^infty left(a_n+a_{n+2}-2a_{n+1}right)frac{1-cosleft(n+1right)x}{4sin^2 frac{x}{2}}$$
which is a positive-term series.
There exists a constant $C>0$, for $0<x<frac{1}{n}$, $$frac{1-cos nx}{4sin^2 frac{x}{2}}ge Cn^2$$
Therefore $$Sleft(xright)ge Csum_{n=1}^{left[frac{1}{x}right]-1} left(n+1right)^2 left(a_n+a_{n+2}-2a_{n+1}right)$$
Then we need to prove that RHS is divergent.
We need a lemma: If $b_n$ is decreasing and tends to $0$, $sum n^k b_n$ diverges, then $sum n^{k+1}left(b_n-b_{n+1}right)$ also diverges.
Since $$Delta n^{k+1}=n^{k+1}-left(n-1right)^{k+1}sim left(k+1right) n^k$$ we know that $sum Delta n^{k+1},b_n$ diverges.
For $M>0$, There exists $N_0$ s.t. $$sum_{n=1}^{N_0}Delta n^{k+1},b_n>M+1$$
As $b_nrightarrow 0$, there exists $N_1$, for all $N>N_1$ we have $N_0^{k+1}b_N<1$.
Then for $N>maxleft{N_0,N_1right}$,$$begin{aligned}sum_{n=1}^N n^{k+1}left(b_n-b_{n+1}right)&=sum_{n=1}^N Delta n^{k+1}left(b_n-b_{N+1}right)\&gesum_{n=1}^{N_0}Delta n^{k+1} left(b_n-b_{N+1}right)\&=sum_{n=1}^{N_0} Delta n^{k+1},b_n -N_0^{k+1} b_{N+1}\&>M+1-1=Mend{aligned}$$
Hence by the lemma we can get $$sum a_n =+infty Rightarrow sum n left(a_n-a_{n+1}right) =+infty Rightarrow sum n^2 left(a_n+a_{n+2}-2a_{n+1}right)=+infty$$
which finally proves the theorem. $blacksquare $
The proof of Thm.3 needs two lemmas as follow.
Lemma 1. If $b_n$ is decreasing and $nb_nrightarrow 0$, then $sum_{n=1}^infty b_n sin nx $ converges uniformly on $mathbb{R}$.
Lemma 2. $S$ has an antiderivative $sum _{n=1}^infty frac{a_n}{n}sin nx$ and $$lim_{deltarightarrow 0^+}int _delta ^pi 2Sleft(xright) cos mx dx=pi a_m$$
The proof of Lemma.1 is a little complex but not difficult, and is taken from an exercise book in mathematical analysis. We will just assume it right.
Proof of Lemma.2:
Notice that the series of $S$ converges uniformly in $left[delta,piright]$, hence we can integrate by terms $$begin{aligned}int_delta^pi 2Sleft(xright)cos mx,dx&=sum_{n=1}^infty int_delta^pi 2a_n cos nx cos mx,dx\&=sum_{n=1}^infty frac{a_n}{n+m}sinleft(n+mright)delta +sum_{nne m}frac{a_n}{n-m}sin left(n-mright)delta +a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)\&overset{def}{=}Theta_1left(deltaright)+Theta_2left(deltaright)+a_mleft(pi-delta+frac{sin 2mdelta}{2m}right)end{aligned}$$
Clearly by lemma 1, $Theta_1$ and $Theta _2$ both converge uniformly on $mathbb{R}$, hence continuous at 0.
So we have $$lim_{deltarightarrow 0^+}int_delta^pi 2Sleft(xright)cos mx,dx=pi a_m$$
Letting $m=0$ we can get that $$int_x^pi Sleft(tright) dt=sum_{n=1}^infty frac{a_n}{n}sin nx$$
By the continuity of $S$ we finish the proof. $blacksquare$
Back to the proof of Thm.3
Note that $S$ is not integrable, that means $$int_0^pi left|Sleft(tright)right| dt= +infty$$
But the improper integral of $S$ converges to $0$ as discussed above.
That means S must have infinitely many positive parts and negative parts near $0$, by continuity there must be infinitely many zeroes. $blacksquare$
So if we are going to find $S$ such that $lim_{xrightarrow 0}Sleft(xright)ne +infty$, we can just find an unintegrable $S$, and the coefficient $a_n$ must not be convex.
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
edited Feb 1 at 20:06
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
answered Jan 28 at 15:12
AntimoniusAntimonius
398110
398110
$begingroup$
Would $S’(x)$ give some useful information?
$endgroup$
– Szeto
Jan 28 at 23:20
$begingroup$
@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
$endgroup$
– Antimonius
Jan 29 at 3:48
add a comment |
$begingroup$
Would $S’(x)$ give some useful information?
$endgroup$
– Szeto
Jan 28 at 23:20
$begingroup$
@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
$endgroup$
– Antimonius
Jan 29 at 3:48
$begingroup$
Would $S’(x)$ give some useful information?
$endgroup$
– Szeto
Jan 28 at 23:20
$begingroup$
Would $S’(x)$ give some useful information?
$endgroup$
– Szeto
Jan 28 at 23:20
$begingroup$
@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
$endgroup$
– Antimonius
Jan 29 at 3:48
$begingroup$
@Szeto But $S$ is not necessarily differentiable, even nowhere differentiable.
$endgroup$
– Antimonius
Jan 29 at 3:48
add a comment |
$begingroup$
All I can think of at this moment is the following very restricted partial result:
Claim. If $(a_n)$ is non-negative, non-increasing and $n a_n leq C$ for some constant $C > 0$, then
$$ lim_{ntoinfty} sum_{n=0}^{infty} a_n cos(n x) = sum_{n=0}^{infty} a_n, $$
regradless of the convergence of $sum_{n=0}^{infty} a_n$.
Since the claim is obvious if $sum_{n=0}^{infty} a_n < infty$, we may focus on the case $sum_{n=0}^{infty} a_n = infty$. To this end, define an auxiliary sequence $(b_n)$ by $a_n - a_{n+1} = b_n$. Then
$b_n geq 0$ and $a_m = a_n + sum_{k=m}^{n-1} b_k$ for $0 leq m < n$,
$a_n = sum_{k=n}^{infty} b_k$, and
$sum_{k=0}^{infty} (k+1)b_k = sum_{n=0}^{infty} a_n = infty$. (This follows from Tonelli's theorem.)
Moreover, we can apply summation by parts
begin{align*}
sum_{n=0}^{N} a_n cos(nx)
&= a_N sum_{n=0}^{N} cos(nx) + sum_{n=0}^{N-1} left( sum_{k=n}^{N-1} b_k right) cos(nx) \
&= a_N D_N(x) + sum_{k=0}^{N-1} b_k D_k(x),
end{align*}
where $D_k(x) = sum_{n=0}^{k} cos(nx)$. Now if $x in (0, pi]$, then $D_n(x)$ is bounded in $n$. So, as $Ntoinfty$, the above converges to
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
= sum_{k=0}^{infty} b_k D_k(x).
end{align*}
Now notice that $D_k(x) = cos(kx/2)sin((k+1)x/2)/sin(x/2)$. So, if $x in (0, 1)$ and $k leq 1/x$, then
$$ D_k(x) geq frac{cos(1/2) cdot frac{2}{pi} (k+1)x/2}{x/2} = c(k+1) $$
for $c = frac{2}{pi}cos(1/2) > 0$. So
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
&geq sum_{0 leq k leq 1/x} b_k D_k(x) - sum_{k > 1/x} frac{b_k}{sin(x/2)} \
&geq c sum_{0 leq k leq 1/x} (k+1)b_k - frac{a_{lceil 1/x rceil}}{sin(x/2)}.
end{align*}
Since $a_{lceil 1/x rceil} = mathcal{O}(x)$ as $x to 0^+$, we have $frac{a_{lceil 1/x rceil}}{sin(x/2)} = mathcal{O}(1)$, and so, letting $xto 0^+$ proves the desired claim.
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
$endgroup$
$begingroup$
Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
$endgroup$
– Antimonius
Feb 1 at 3:18
add a comment |
$begingroup$
All I can think of at this moment is the following very restricted partial result:
Claim. If $(a_n)$ is non-negative, non-increasing and $n a_n leq C$ for some constant $C > 0$, then
$$ lim_{ntoinfty} sum_{n=0}^{infty} a_n cos(n x) = sum_{n=0}^{infty} a_n, $$
regradless of the convergence of $sum_{n=0}^{infty} a_n$.
Since the claim is obvious if $sum_{n=0}^{infty} a_n < infty$, we may focus on the case $sum_{n=0}^{infty} a_n = infty$. To this end, define an auxiliary sequence $(b_n)$ by $a_n - a_{n+1} = b_n$. Then
$b_n geq 0$ and $a_m = a_n + sum_{k=m}^{n-1} b_k$ for $0 leq m < n$,
$a_n = sum_{k=n}^{infty} b_k$, and
$sum_{k=0}^{infty} (k+1)b_k = sum_{n=0}^{infty} a_n = infty$. (This follows from Tonelli's theorem.)
Moreover, we can apply summation by parts
begin{align*}
sum_{n=0}^{N} a_n cos(nx)
&= a_N sum_{n=0}^{N} cos(nx) + sum_{n=0}^{N-1} left( sum_{k=n}^{N-1} b_k right) cos(nx) \
&= a_N D_N(x) + sum_{k=0}^{N-1} b_k D_k(x),
end{align*}
where $D_k(x) = sum_{n=0}^{k} cos(nx)$. Now if $x in (0, pi]$, then $D_n(x)$ is bounded in $n$. So, as $Ntoinfty$, the above converges to
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
= sum_{k=0}^{infty} b_k D_k(x).
end{align*}
Now notice that $D_k(x) = cos(kx/2)sin((k+1)x/2)/sin(x/2)$. So, if $x in (0, 1)$ and $k leq 1/x$, then
$$ D_k(x) geq frac{cos(1/2) cdot frac{2}{pi} (k+1)x/2}{x/2} = c(k+1) $$
for $c = frac{2}{pi}cos(1/2) > 0$. So
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
&geq sum_{0 leq k leq 1/x} b_k D_k(x) - sum_{k > 1/x} frac{b_k}{sin(x/2)} \
&geq c sum_{0 leq k leq 1/x} (k+1)b_k - frac{a_{lceil 1/x rceil}}{sin(x/2)}.
end{align*}
Since $a_{lceil 1/x rceil} = mathcal{O}(x)$ as $x to 0^+$, we have $frac{a_{lceil 1/x rceil}}{sin(x/2)} = mathcal{O}(1)$, and so, letting $xto 0^+$ proves the desired claim.
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
$endgroup$
$begingroup$
Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
$endgroup$
– Antimonius
Feb 1 at 3:18
add a comment |
$begingroup$
All I can think of at this moment is the following very restricted partial result:
Claim. If $(a_n)$ is non-negative, non-increasing and $n a_n leq C$ for some constant $C > 0$, then
$$ lim_{ntoinfty} sum_{n=0}^{infty} a_n cos(n x) = sum_{n=0}^{infty} a_n, $$
regradless of the convergence of $sum_{n=0}^{infty} a_n$.
Since the claim is obvious if $sum_{n=0}^{infty} a_n < infty$, we may focus on the case $sum_{n=0}^{infty} a_n = infty$. To this end, define an auxiliary sequence $(b_n)$ by $a_n - a_{n+1} = b_n$. Then
$b_n geq 0$ and $a_m = a_n + sum_{k=m}^{n-1} b_k$ for $0 leq m < n$,
$a_n = sum_{k=n}^{infty} b_k$, and
$sum_{k=0}^{infty} (k+1)b_k = sum_{n=0}^{infty} a_n = infty$. (This follows from Tonelli's theorem.)
Moreover, we can apply summation by parts
begin{align*}
sum_{n=0}^{N} a_n cos(nx)
&= a_N sum_{n=0}^{N} cos(nx) + sum_{n=0}^{N-1} left( sum_{k=n}^{N-1} b_k right) cos(nx) \
&= a_N D_N(x) + sum_{k=0}^{N-1} b_k D_k(x),
end{align*}
where $D_k(x) = sum_{n=0}^{k} cos(nx)$. Now if $x in (0, pi]$, then $D_n(x)$ is bounded in $n$. So, as $Ntoinfty$, the above converges to
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
= sum_{k=0}^{infty} b_k D_k(x).
end{align*}
Now notice that $D_k(x) = cos(kx/2)sin((k+1)x/2)/sin(x/2)$. So, if $x in (0, 1)$ and $k leq 1/x$, then
$$ D_k(x) geq frac{cos(1/2) cdot frac{2}{pi} (k+1)x/2}{x/2} = c(k+1) $$
for $c = frac{2}{pi}cos(1/2) > 0$. So
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
&geq sum_{0 leq k leq 1/x} b_k D_k(x) - sum_{k > 1/x} frac{b_k}{sin(x/2)} \
&geq c sum_{0 leq k leq 1/x} (k+1)b_k - frac{a_{lceil 1/x rceil}}{sin(x/2)}.
end{align*}
Since $a_{lceil 1/x rceil} = mathcal{O}(x)$ as $x to 0^+$, we have $frac{a_{lceil 1/x rceil}}{sin(x/2)} = mathcal{O}(1)$, and so, letting $xto 0^+$ proves the desired claim.
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
$endgroup$
All I can think of at this moment is the following very restricted partial result:
Claim. If $(a_n)$ is non-negative, non-increasing and $n a_n leq C$ for some constant $C > 0$, then
$$ lim_{ntoinfty} sum_{n=0}^{infty} a_n cos(n x) = sum_{n=0}^{infty} a_n, $$
regradless of the convergence of $sum_{n=0}^{infty} a_n$.
Since the claim is obvious if $sum_{n=0}^{infty} a_n < infty$, we may focus on the case $sum_{n=0}^{infty} a_n = infty$. To this end, define an auxiliary sequence $(b_n)$ by $a_n - a_{n+1} = b_n$. Then
$b_n geq 0$ and $a_m = a_n + sum_{k=m}^{n-1} b_k$ for $0 leq m < n$,
$a_n = sum_{k=n}^{infty} b_k$, and
$sum_{k=0}^{infty} (k+1)b_k = sum_{n=0}^{infty} a_n = infty$. (This follows from Tonelli's theorem.)
Moreover, we can apply summation by parts
begin{align*}
sum_{n=0}^{N} a_n cos(nx)
&= a_N sum_{n=0}^{N} cos(nx) + sum_{n=0}^{N-1} left( sum_{k=n}^{N-1} b_k right) cos(nx) \
&= a_N D_N(x) + sum_{k=0}^{N-1} b_k D_k(x),
end{align*}
where $D_k(x) = sum_{n=0}^{k} cos(nx)$. Now if $x in (0, pi]$, then $D_n(x)$ is bounded in $n$. So, as $Ntoinfty$, the above converges to
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
= sum_{k=0}^{infty} b_k D_k(x).
end{align*}
Now notice that $D_k(x) = cos(kx/2)sin((k+1)x/2)/sin(x/2)$. So, if $x in (0, 1)$ and $k leq 1/x$, then
$$ D_k(x) geq frac{cos(1/2) cdot frac{2}{pi} (k+1)x/2}{x/2} = c(k+1) $$
for $c = frac{2}{pi}cos(1/2) > 0$. So
begin{align*}
sum_{n=0}^{infty} a_n cos(nx)
&geq sum_{0 leq k leq 1/x} b_k D_k(x) - sum_{k > 1/x} frac{b_k}{sin(x/2)} \
&geq c sum_{0 leq k leq 1/x} (k+1)b_k - frac{a_{lceil 1/x rceil}}{sin(x/2)}.
end{align*}
Since $a_{lceil 1/x rceil} = mathcal{O}(x)$ as $x to 0^+$, we have $frac{a_{lceil 1/x rceil}}{sin(x/2)} = mathcal{O}(1)$, and so, letting $xto 0^+$ proves the desired claim.
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
answered Jan 29 at 9:47
Sangchul LeeSangchul Lee
96.1k12171281
96.1k12171281
$begingroup$
Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
$endgroup$
– Antimonius
Feb 1 at 3:18
add a comment |
$begingroup$
Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
$endgroup$
– Antimonius
Feb 1 at 3:18
$begingroup$
Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
$endgroup$
– Antimonius
Feb 1 at 3:18
$begingroup$
Nice answer. It can also explain an interesting phenomenon that $Sleft(xright)$ diverges to $infty$ at the same rate of $sum_{n=1}^{left[frac{1}{x}right]}a_n$ under certain condition. And recently I proved some other little theorems of the divergence to $infty$ of $Sleft(xright)$.
$endgroup$
– Antimonius
Feb 1 at 3:18
add a comment |
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