Mixing
Closed form expression for the harmonic sum $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2cdot4^n}{2n choose n}$
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I'm wondering if one could derive a closed form expression for the series
$$sum_{n=1}^{infty}frac{H_{2n}}{n^2cdot4^n}{2n choose n}$$
$$text{With } text{ } text{ } text{ }H_n=sum_{k=1}^{n}frac{1}{k}text{ } text{ } text{} text{ } text{ }text{the } n^{th} text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
real-analysis calculus sequences-and-series summation
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
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add a comment |
$begingroup$
I'm wondering if one could derive a closed form expression for the series
$$sum_{n=1}^{infty}frac{H_{2n}}{n^2cdot4^n}{2n choose n}$$
$$text{With } text{ } text{ } text{ }H_n=sum_{k=1}^{n}frac{1}{k}text{ } text{ } text{} text{ } text{ }text{the } n^{th} text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
real-analysis calculus sequences-and-series summation
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
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$begingroup$
Don't you mean $H_n=sum_{k=1}^{n}frac{1}{k}$?
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– James Arathoon
Jan 24 at 1:13
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Oh yes sorry, now it's corrected !
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– Harmonic Sun
Jan 24 at 1:14
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Would the fact that $frac{{2nchoose n}}{4^n}=frac{1}{2pi}int_0^{2pi} cos^{2n} x,dx$ be useful if you can switch integration and summation?
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– Zachary
Jan 24 at 1:47
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Looks like a very good idea, but as far as I know there is no known closed form for the power series $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function...
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– Harmonic Sun
Jan 24 at 2:32
add a comment |
$begingroup$
I'm wondering if one could derive a closed form expression for the series
$$sum_{n=1}^{infty}frac{H_{2n}}{n^2cdot4^n}{2n choose n}$$
$$text{With } text{ } text{ } text{ }H_n=sum_{k=1}^{n}frac{1}{k}text{ } text{ } text{} text{ } text{ }text{the } n^{th} text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
real-analysis calculus sequences-and-series summation
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
$endgroup$
I'm wondering if one could derive a closed form expression for the series
$$sum_{n=1}^{infty}frac{H_{2n}}{n^2cdot4^n}{2n choose n}$$
$$text{With } text{ } text{ } text{ }H_n=sum_{k=1}^{n}frac{1}{k}text{ } text{ } text{} text{ } text{ }text{the } n^{th} text{ harmonic number.}$$
Now, I know series involving harmonic numbers are well suited for a summation by part (or Abel's transformation) approach, but it doesn't lead anywere here, at least not in this state.
Any suggestions ?
real-analysis calculus sequences-and-series summation
real-analysis calculus sequences-and-series summation
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
edited Jan 24 at 1:14
Harmonic Sun
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
asked Jan 24 at 1:02
Harmonic SunHarmonic Sun
65210
65210
$begingroup$
Don't you mean $H_n=sum_{k=1}^{n}frac{1}{k}$?
$endgroup$
– James Arathoon
Jan 24 at 1:13
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Oh yes sorry, now it's corrected !
$endgroup$
– Harmonic Sun
Jan 24 at 1:14
$begingroup$
Would the fact that $frac{{2nchoose n}}{4^n}=frac{1}{2pi}int_0^{2pi} cos^{2n} x,dx$ be useful if you can switch integration and summation?
$endgroup$
– Zachary
Jan 24 at 1:47
$begingroup$
Looks like a very good idea, but as far as I know there is no known closed form for the power series $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function...
$endgroup$
– Harmonic Sun
Jan 24 at 2:32
add a comment |
$begingroup$
Don't you mean $H_n=sum_{k=1}^{n}frac{1}{k}$?
$endgroup$
– James Arathoon
Jan 24 at 1:13
$begingroup$
Oh yes sorry, now it's corrected !
$endgroup$
– Harmonic Sun
Jan 24 at 1:14
$begingroup$
Would the fact that $frac{{2nchoose n}}{4^n}=frac{1}{2pi}int_0^{2pi} cos^{2n} x,dx$ be useful if you can switch integration and summation?
$endgroup$
– Zachary
Jan 24 at 1:47
$begingroup$
Looks like a very good idea, but as far as I know there is no known closed form for the power series $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function...
$endgroup$
– Harmonic Sun
Jan 24 at 2:32
$begingroup$
Don't you mean $H_n=sum_{k=1}^{n}frac{1}{k}$?
$endgroup$
– James Arathoon
Jan 24 at 1:13
$begingroup$
Don't you mean $H_n=sum_{k=1}^{n}frac{1}{k}$?
$endgroup$
– James Arathoon
Jan 24 at 1:13
$begingroup$
Oh yes sorry, now it's corrected !
$endgroup$
– Harmonic Sun
Jan 24 at 1:14
$begingroup$
Oh yes sorry, now it's corrected !
$endgroup$
– Harmonic Sun
Jan 24 at 1:14
$begingroup$
Would the fact that $frac{{2nchoose n}}{4^n}=frac{1}{2pi}int_0^{2pi} cos^{2n} x,dx$ be useful if you can switch integration and summation?
$endgroup$
– Zachary
Jan 24 at 1:47
$begingroup$
Would the fact that $frac{{2nchoose n}}{4^n}=frac{1}{2pi}int_0^{2pi} cos^{2n} x,dx$ be useful if you can switch integration and summation?
$endgroup$
– Zachary
Jan 24 at 1:47
$begingroup$
Looks like a very good idea, but as far as I know there is no known closed form for the power series $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function...
$endgroup$
– Harmonic Sun
Jan 24 at 2:32
$begingroup$
Looks like a very good idea, but as far as I know there is no known closed form for the power series $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function...
$endgroup$
– Harmonic Sun
Jan 24 at 2:32
add a comment |
2 Answers
2
active
oldest
votes
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For $x in [0,1]$ let
$$ f(x) = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} x^{2n} , . $$
Using the power series of $arcsin$ we find
$$ x frac{mathrm{d}}{mathrm{d} x} x frac{mathrm{d}}{mathrm{d} x} f(x) = 4 frac{mathrm{d}}{mathrm{d} x} [arcsin(x) - x] = 4 left[frac{1}{sqrt{1-x^2}} - 1 right] $$
for $x in [0,1)$ . In particular,
$$ f'(1) = 4 int limits_0^1 frac{1}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x stackrel{x=sqrt{1-y^2}}{=} 4 int limits_0^1 frac{mathrm{d} y}{1+y} = 4 ln(2) , . $$
Now we can compute
begin{align}
S &equiv sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} int limits_0^1 frac{1-x^{2n}}{1-x} , mathrm{d} x = int limits_0^1 frac{f(1) - f(x)}{1-x} , mathrm{d} x \
&= int limits_0^1 frac{- ln(1-x)}{x} x f'(x) , mathrm{d} x
= operatorname{Li}_2 (1) f'(1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x \
&= operatorname{Li}_2 (1) f'(1) + 4 operatorname{Li}_3 (1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x sqrt{1-x^2}} , mathrm{d} x equiv 4 left[frac{pi^2}{6} ln(2) + zeta(3) - Iright] , .
end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
begin{align}
I &= int limits_0^infty t int limits_0^1 frac{mathrm{d} x}{(mathrm{e}^t - x) sqrt{1-x^2}} , mathrm{d} t stackrel{(*)}{=} int limits_0^infty frac{t left[frac{pi}{2} + arcsin(mathrm{e}^{-t})right]}{sqrt{mathrm{e}^{2t}-1}} , mathrm{d} t \
&stackrel{mathrm{e}^{-t} = sin(u)}{=} frac{1}{2} int limits_0^{pi/2} -ln[sin(u)] (pi + 2 u) , mathrm{d} u = frac{1}{2} int limits_0^{pi/2} u (pi + u) cot(u) , mathrm{d} u \
&= frac{1}{2} [pi K_1^{(1)} + K_2^{(1)}] = frac{3}{8}pi^2 ln(2) - frac{7}{16} zeta(3) , .
end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ boxed{S = sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = frac{23}{4} zeta(3) - frac{5}{6} pi^2 ln(2)} , . $$
Proof of $(*)$:
For $a in [0,1]$ let
$$ g(a) = int limits_0^1 frac{-ln(1-a x)}{x sqrt{1-x^2}} , mathrm{d} x= sum limits_{n=1}^infty frac{a^n}{n} int limits_0^{pi/2} sin^{n-1} (t) , mathrm{d} t , .$$
Using Wallis' integrals we find
$$ g(a) = frac{pi}{2} sum limits_{k=0}^infty frac{{2k choose k} a^{2k+1}}{4^k(2k+1)} + frac{1}{4} sum limits_{m=1}^infty frac{4^k a^{2k}}{k^2 {2k choose k}} = frac{pi}{2} arcsin(a) + frac{1}{2} arcsin^2 (a) , . $$
Therefore
$$ int limits_0^1 frac{mathrm{d} x}{(1-a x)sqrt{1-x^2}} = g'(a) = frac{frac{pi}{2} + arcsin{a}}{sqrt{1-a^2}} $$
holds for $a in [0,1)$ .
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
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1
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This is a nice answer. Now, when I look at my junk, I am just ashamed.
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– Claude Leibovici
Jan 24 at 5:44
1
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Yes indeed, this is a very nice answer.
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– omegadot
Jan 24 at 5:46
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Thank you very much, this is a great answer !
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– Harmonic Sun
Jan 24 at 15:03
add a comment |
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This is not an answer but it is too long for a comment.
Considering
$$a_n=frac{H_{2n}}{n^2,4^n}{2n choose n}qquad text{and} qquad S_p=sum_{n=1}^p a_n$$ none of the CAS I tried was able to find an expression for the partial sums or the infinite sum. Numerically, as shown below, the convergence looks to be extremely slow
$$left(
begin{array}{cc}
p & S_p \
1000 & 1.21081501745 \
2000 & 1.21088004598 \
3000 & 1.21089738494 \
4000 & 1.21090493158 \
5000 & 1.21090901996 \
6000 & 1.21091153294 \
7000 & 1.21091321066 \
8000 & 1.21091439815 \
9000 & 1.21091527609 \
10000 & 1.21091594745
end{array}
right)$$ which can be explained by the fact that, for large values of $n$
$$frac {a_{n+1}} {a_n} simeq 1+frac{2-5( log (2n)+ gamma) }{2 n left(log(2n)+gamma right)}$$ For the infinite summation, the result seems to be close to $1.2109201$ which is not identified by inverse symbolic calculators.
For large values of $n$, we also can find
$$a_nsimeq b_n=frac{ log (2n)+gamma }{n^{5/2},sqrt{pi }}$$ which does not help much even if
$$sum_{n=1}^infty b_n=frac{(gamma+log(2)) zeta left(frac{5}{2}right)-zeta 'left(frac{5}{2}right)}{sqrt{pi }}approx 1.18001$$
However, numerically, this can be of some help writing
$$S_infty=S_p+sum_{n=p+1}^infty b_n$$
$$left(
begin{array}{cc}
p & S_infty approx \
100 & 1.2109213325 \
200 & 1.2109203863 \
300 & 1.2109202368 \
400 & 1.2109201900 \
500 & 1.2109201700 \
600 & 1.2109201590 \
700 & 1.2109201535 \
800 & 1.2109201498 \
900 & 1.2109201475 \
1000 & 1.2109201458
end{array}
right)$$
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
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First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
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– James Arathoon
Jan 24 at 12:39
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or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
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– James Arathoon
Jan 24 at 13:20
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First comment should read after partial term 10
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– James Arathoon
Jan 24 at 13:27
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@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
$endgroup$
– Claude Leibovici
Jan 24 at 16:00
add a comment |
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2 Answers
2
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
For $x in [0,1]$ let
$$ f(x) = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} x^{2n} , . $$
Using the power series of $arcsin$ we find
$$ x frac{mathrm{d}}{mathrm{d} x} x frac{mathrm{d}}{mathrm{d} x} f(x) = 4 frac{mathrm{d}}{mathrm{d} x} [arcsin(x) - x] = 4 left[frac{1}{sqrt{1-x^2}} - 1 right] $$
for $x in [0,1)$ . In particular,
$$ f'(1) = 4 int limits_0^1 frac{1}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x stackrel{x=sqrt{1-y^2}}{=} 4 int limits_0^1 frac{mathrm{d} y}{1+y} = 4 ln(2) , . $$
Now we can compute
begin{align}
S &equiv sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} int limits_0^1 frac{1-x^{2n}}{1-x} , mathrm{d} x = int limits_0^1 frac{f(1) - f(x)}{1-x} , mathrm{d} x \
&= int limits_0^1 frac{- ln(1-x)}{x} x f'(x) , mathrm{d} x
= operatorname{Li}_2 (1) f'(1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x \
&= operatorname{Li}_2 (1) f'(1) + 4 operatorname{Li}_3 (1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x sqrt{1-x^2}} , mathrm{d} x equiv 4 left[frac{pi^2}{6} ln(2) + zeta(3) - Iright] , .
end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
begin{align}
I &= int limits_0^infty t int limits_0^1 frac{mathrm{d} x}{(mathrm{e}^t - x) sqrt{1-x^2}} , mathrm{d} t stackrel{(*)}{=} int limits_0^infty frac{t left[frac{pi}{2} + arcsin(mathrm{e}^{-t})right]}{sqrt{mathrm{e}^{2t}-1}} , mathrm{d} t \
&stackrel{mathrm{e}^{-t} = sin(u)}{=} frac{1}{2} int limits_0^{pi/2} -ln[sin(u)] (pi + 2 u) , mathrm{d} u = frac{1}{2} int limits_0^{pi/2} u (pi + u) cot(u) , mathrm{d} u \
&= frac{1}{2} [pi K_1^{(1)} + K_2^{(1)}] = frac{3}{8}pi^2 ln(2) - frac{7}{16} zeta(3) , .
end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ boxed{S = sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = frac{23}{4} zeta(3) - frac{5}{6} pi^2 ln(2)} , . $$
Proof of $(*)$:
For $a in [0,1]$ let
$$ g(a) = int limits_0^1 frac{-ln(1-a x)}{x sqrt{1-x^2}} , mathrm{d} x= sum limits_{n=1}^infty frac{a^n}{n} int limits_0^{pi/2} sin^{n-1} (t) , mathrm{d} t , .$$
Using Wallis' integrals we find
$$ g(a) = frac{pi}{2} sum limits_{k=0}^infty frac{{2k choose k} a^{2k+1}}{4^k(2k+1)} + frac{1}{4} sum limits_{m=1}^infty frac{4^k a^{2k}}{k^2 {2k choose k}} = frac{pi}{2} arcsin(a) + frac{1}{2} arcsin^2 (a) , . $$
Therefore
$$ int limits_0^1 frac{mathrm{d} x}{(1-a x)sqrt{1-x^2}} = g'(a) = frac{frac{pi}{2} + arcsin{a}}{sqrt{1-a^2}} $$
holds for $a in [0,1)$ .
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
$endgroup$
1
$begingroup$
This is a nice answer. Now, when I look at my junk, I am just ashamed.
$endgroup$
– Claude Leibovici
Jan 24 at 5:44
1
$begingroup$
Yes indeed, this is a very nice answer.
$endgroup$
– omegadot
Jan 24 at 5:46
$begingroup$
Thank you very much, this is a great answer !
$endgroup$
– Harmonic Sun
Jan 24 at 15:03
add a comment |
$begingroup$
For $x in [0,1]$ let
$$ f(x) = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} x^{2n} , . $$
Using the power series of $arcsin$ we find
$$ x frac{mathrm{d}}{mathrm{d} x} x frac{mathrm{d}}{mathrm{d} x} f(x) = 4 frac{mathrm{d}}{mathrm{d} x} [arcsin(x) - x] = 4 left[frac{1}{sqrt{1-x^2}} - 1 right] $$
for $x in [0,1)$ . In particular,
$$ f'(1) = 4 int limits_0^1 frac{1}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x stackrel{x=sqrt{1-y^2}}{=} 4 int limits_0^1 frac{mathrm{d} y}{1+y} = 4 ln(2) , . $$
Now we can compute
begin{align}
S &equiv sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} int limits_0^1 frac{1-x^{2n}}{1-x} , mathrm{d} x = int limits_0^1 frac{f(1) - f(x)}{1-x} , mathrm{d} x \
&= int limits_0^1 frac{- ln(1-x)}{x} x f'(x) , mathrm{d} x
= operatorname{Li}_2 (1) f'(1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x \
&= operatorname{Li}_2 (1) f'(1) + 4 operatorname{Li}_3 (1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x sqrt{1-x^2}} , mathrm{d} x equiv 4 left[frac{pi^2}{6} ln(2) + zeta(3) - Iright] , .
end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
begin{align}
I &= int limits_0^infty t int limits_0^1 frac{mathrm{d} x}{(mathrm{e}^t - x) sqrt{1-x^2}} , mathrm{d} t stackrel{(*)}{=} int limits_0^infty frac{t left[frac{pi}{2} + arcsin(mathrm{e}^{-t})right]}{sqrt{mathrm{e}^{2t}-1}} , mathrm{d} t \
&stackrel{mathrm{e}^{-t} = sin(u)}{=} frac{1}{2} int limits_0^{pi/2} -ln[sin(u)] (pi + 2 u) , mathrm{d} u = frac{1}{2} int limits_0^{pi/2} u (pi + u) cot(u) , mathrm{d} u \
&= frac{1}{2} [pi K_1^{(1)} + K_2^{(1)}] = frac{3}{8}pi^2 ln(2) - frac{7}{16} zeta(3) , .
end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ boxed{S = sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = frac{23}{4} zeta(3) - frac{5}{6} pi^2 ln(2)} , . $$
Proof of $(*)$:
For $a in [0,1]$ let
$$ g(a) = int limits_0^1 frac{-ln(1-a x)}{x sqrt{1-x^2}} , mathrm{d} x= sum limits_{n=1}^infty frac{a^n}{n} int limits_0^{pi/2} sin^{n-1} (t) , mathrm{d} t , .$$
Using Wallis' integrals we find
$$ g(a) = frac{pi}{2} sum limits_{k=0}^infty frac{{2k choose k} a^{2k+1}}{4^k(2k+1)} + frac{1}{4} sum limits_{m=1}^infty frac{4^k a^{2k}}{k^2 {2k choose k}} = frac{pi}{2} arcsin(a) + frac{1}{2} arcsin^2 (a) , . $$
Therefore
$$ int limits_0^1 frac{mathrm{d} x}{(1-a x)sqrt{1-x^2}} = g'(a) = frac{frac{pi}{2} + arcsin{a}}{sqrt{1-a^2}} $$
holds for $a in [0,1)$ .
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
$endgroup$
1
$begingroup$
This is a nice answer. Now, when I look at my junk, I am just ashamed.
$endgroup$
– Claude Leibovici
Jan 24 at 5:44
1
$begingroup$
Yes indeed, this is a very nice answer.
$endgroup$
– omegadot
Jan 24 at 5:46
$begingroup$
Thank you very much, this is a great answer !
$endgroup$
– Harmonic Sun
Jan 24 at 15:03
add a comment |
$begingroup$
For $x in [0,1]$ let
$$ f(x) = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} x^{2n} , . $$
Using the power series of $arcsin$ we find
$$ x frac{mathrm{d}}{mathrm{d} x} x frac{mathrm{d}}{mathrm{d} x} f(x) = 4 frac{mathrm{d}}{mathrm{d} x} [arcsin(x) - x] = 4 left[frac{1}{sqrt{1-x^2}} - 1 right] $$
for $x in [0,1)$ . In particular,
$$ f'(1) = 4 int limits_0^1 frac{1}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x stackrel{x=sqrt{1-y^2}}{=} 4 int limits_0^1 frac{mathrm{d} y}{1+y} = 4 ln(2) , . $$
Now we can compute
begin{align}
S &equiv sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} int limits_0^1 frac{1-x^{2n}}{1-x} , mathrm{d} x = int limits_0^1 frac{f(1) - f(x)}{1-x} , mathrm{d} x \
&= int limits_0^1 frac{- ln(1-x)}{x} x f'(x) , mathrm{d} x
= operatorname{Li}_2 (1) f'(1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x \
&= operatorname{Li}_2 (1) f'(1) + 4 operatorname{Li}_3 (1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x sqrt{1-x^2}} , mathrm{d} x equiv 4 left[frac{pi^2}{6} ln(2) + zeta(3) - Iright] , .
end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
begin{align}
I &= int limits_0^infty t int limits_0^1 frac{mathrm{d} x}{(mathrm{e}^t - x) sqrt{1-x^2}} , mathrm{d} t stackrel{(*)}{=} int limits_0^infty frac{t left[frac{pi}{2} + arcsin(mathrm{e}^{-t})right]}{sqrt{mathrm{e}^{2t}-1}} , mathrm{d} t \
&stackrel{mathrm{e}^{-t} = sin(u)}{=} frac{1}{2} int limits_0^{pi/2} -ln[sin(u)] (pi + 2 u) , mathrm{d} u = frac{1}{2} int limits_0^{pi/2} u (pi + u) cot(u) , mathrm{d} u \
&= frac{1}{2} [pi K_1^{(1)} + K_2^{(1)}] = frac{3}{8}pi^2 ln(2) - frac{7}{16} zeta(3) , .
end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ boxed{S = sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = frac{23}{4} zeta(3) - frac{5}{6} pi^2 ln(2)} , . $$
Proof of $(*)$:
For $a in [0,1]$ let
$$ g(a) = int limits_0^1 frac{-ln(1-a x)}{x sqrt{1-x^2}} , mathrm{d} x= sum limits_{n=1}^infty frac{a^n}{n} int limits_0^{pi/2} sin^{n-1} (t) , mathrm{d} t , .$$
Using Wallis' integrals we find
$$ g(a) = frac{pi}{2} sum limits_{k=0}^infty frac{{2k choose k} a^{2k+1}}{4^k(2k+1)} + frac{1}{4} sum limits_{m=1}^infty frac{4^k a^{2k}}{k^2 {2k choose k}} = frac{pi}{2} arcsin(a) + frac{1}{2} arcsin^2 (a) , . $$
Therefore
$$ int limits_0^1 frac{mathrm{d} x}{(1-a x)sqrt{1-x^2}} = g'(a) = frac{frac{pi}{2} + arcsin{a}}{sqrt{1-a^2}} $$
holds for $a in [0,1)$ .
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
$endgroup$
For $x in [0,1]$ let
$$ f(x) = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} x^{2n} , . $$
Using the power series of $arcsin$ we find
$$ x frac{mathrm{d}}{mathrm{d} x} x frac{mathrm{d}}{mathrm{d} x} f(x) = 4 frac{mathrm{d}}{mathrm{d} x} [arcsin(x) - x] = 4 left[frac{1}{sqrt{1-x^2}} - 1 right] $$
for $x in [0,1)$ . In particular,
$$ f'(1) = 4 int limits_0^1 frac{1}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x stackrel{x=sqrt{1-y^2}}{=} 4 int limits_0^1 frac{mathrm{d} y}{1+y} = 4 ln(2) , . $$
Now we can compute
begin{align}
S &equiv sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = sum limits_{n=1}^infty frac{{2n choose n}}{n^2 4^n} int limits_0^1 frac{1-x^{2n}}{1-x} , mathrm{d} x = int limits_0^1 frac{f(1) - f(x)}{1-x} , mathrm{d} x \
&= int limits_0^1 frac{- ln(1-x)}{x} x f'(x) , mathrm{d} x
= operatorname{Li}_2 (1) f'(1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x} left[frac{1}{sqrt{1-x^2}} - 1 right] , mathrm{d} x \
&= operatorname{Li}_2 (1) f'(1) + 4 operatorname{Li}_3 (1) - 4 int limits_0^1 frac{operatorname{Li}_2 (x)}{x sqrt{1-x^2}} , mathrm{d} x equiv 4 left[frac{pi^2}{6} ln(2) + zeta(3) - Iright] , .
end{align}
In order to find $I$ we use a well-known integral representation of the dilogarithm:
begin{align}
I &= int limits_0^infty t int limits_0^1 frac{mathrm{d} x}{(mathrm{e}^t - x) sqrt{1-x^2}} , mathrm{d} t stackrel{(*)}{=} int limits_0^infty frac{t left[frac{pi}{2} + arcsin(mathrm{e}^{-t})right]}{sqrt{mathrm{e}^{2t}-1}} , mathrm{d} t \
&stackrel{mathrm{e}^{-t} = sin(u)}{=} frac{1}{2} int limits_0^{pi/2} -ln[sin(u)] (pi + 2 u) , mathrm{d} u = frac{1}{2} int limits_0^{pi/2} u (pi + u) cot(u) , mathrm{d} u \
&= frac{1}{2} [pi K_1^{(1)} + K_2^{(1)}] = frac{3}{8}pi^2 ln(2) - frac{7}{16} zeta(3) , .
end{align}
The integrals $ K_n^{(m)}$ are discussed in this question. Combining this result and the previous expression for the sum we end up with
$$ boxed{S = sum limits_{n=1}^infty frac{H_{2n} {2n choose n}}{n^2 4^n} = frac{23}{4} zeta(3) - frac{5}{6} pi^2 ln(2)} , . $$
Proof of $(*)$:
For $a in [0,1]$ let
$$ g(a) = int limits_0^1 frac{-ln(1-a x)}{x sqrt{1-x^2}} , mathrm{d} x= sum limits_{n=1}^infty frac{a^n}{n} int limits_0^{pi/2} sin^{n-1} (t) , mathrm{d} t , .$$
Using Wallis' integrals we find
$$ g(a) = frac{pi}{2} sum limits_{k=0}^infty frac{{2k choose k} a^{2k+1}}{4^k(2k+1)} + frac{1}{4} sum limits_{m=1}^infty frac{4^k a^{2k}}{k^2 {2k choose k}} = frac{pi}{2} arcsin(a) + frac{1}{2} arcsin^2 (a) , . $$
Therefore
$$ int limits_0^1 frac{mathrm{d} x}{(1-a x)sqrt{1-x^2}} = g'(a) = frac{frac{pi}{2} + arcsin{a}}{sqrt{1-a^2}} $$
holds for $a in [0,1)$ .
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
edited Jan 24 at 6:00
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
answered Jan 24 at 4:48
ComplexYetTrivialComplexYetTrivial
4,8332631
4,8332631
1
$begingroup$
This is a nice answer. Now, when I look at my junk, I am just ashamed.
$endgroup$
– Claude Leibovici
Jan 24 at 5:44
1
$begingroup$
Yes indeed, this is a very nice answer.
$endgroup$
– omegadot
Jan 24 at 5:46
$begingroup$
Thank you very much, this is a great answer !
$endgroup$
– Harmonic Sun
Jan 24 at 15:03
add a comment |
1
$begingroup$
This is a nice answer. Now, when I look at my junk, I am just ashamed.
$endgroup$
– Claude Leibovici
Jan 24 at 5:44
1
$begingroup$
Yes indeed, this is a very nice answer.
$endgroup$
– omegadot
Jan 24 at 5:46
$begingroup$
Thank you very much, this is a great answer !
$endgroup$
– Harmonic Sun
Jan 24 at 15:03
1
1
$begingroup$
This is a nice answer. Now, when I look at my junk, I am just ashamed.
$endgroup$
– Claude Leibovici
Jan 24 at 5:44
$begingroup$
This is a nice answer. Now, when I look at my junk, I am just ashamed.
$endgroup$
– Claude Leibovici
Jan 24 at 5:44
1
1
$begingroup$
Yes indeed, this is a very nice answer.
$endgroup$
– omegadot
Jan 24 at 5:46
$begingroup$
Yes indeed, this is a very nice answer.
$endgroup$
– omegadot
Jan 24 at 5:46
$begingroup$
Thank you very much, this is a great answer !
$endgroup$
– Harmonic Sun
Jan 24 at 15:03
$begingroup$
Thank you very much, this is a great answer !
$endgroup$
– Harmonic Sun
Jan 24 at 15:03
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
Considering
$$a_n=frac{H_{2n}}{n^2,4^n}{2n choose n}qquad text{and} qquad S_p=sum_{n=1}^p a_n$$ none of the CAS I tried was able to find an expression for the partial sums or the infinite sum. Numerically, as shown below, the convergence looks to be extremely slow
$$left(
begin{array}{cc}
p & S_p \
1000 & 1.21081501745 \
2000 & 1.21088004598 \
3000 & 1.21089738494 \
4000 & 1.21090493158 \
5000 & 1.21090901996 \
6000 & 1.21091153294 \
7000 & 1.21091321066 \
8000 & 1.21091439815 \
9000 & 1.21091527609 \
10000 & 1.21091594745
end{array}
right)$$ which can be explained by the fact that, for large values of $n$
$$frac {a_{n+1}} {a_n} simeq 1+frac{2-5( log (2n)+ gamma) }{2 n left(log(2n)+gamma right)}$$ For the infinite summation, the result seems to be close to $1.2109201$ which is not identified by inverse symbolic calculators.
For large values of $n$, we also can find
$$a_nsimeq b_n=frac{ log (2n)+gamma }{n^{5/2},sqrt{pi }}$$ which does not help much even if
$$sum_{n=1}^infty b_n=frac{(gamma+log(2)) zeta left(frac{5}{2}right)-zeta 'left(frac{5}{2}right)}{sqrt{pi }}approx 1.18001$$
However, numerically, this can be of some help writing
$$S_infty=S_p+sum_{n=p+1}^infty b_n$$
$$left(
begin{array}{cc}
p & S_infty approx \
100 & 1.2109213325 \
200 & 1.2109203863 \
300 & 1.2109202368 \
400 & 1.2109201900 \
500 & 1.2109201700 \
600 & 1.2109201590 \
700 & 1.2109201535 \
800 & 1.2109201498 \
900 & 1.2109201475 \
1000 & 1.2109201458
end{array}
right)$$
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
$begingroup$
First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
$endgroup$
– James Arathoon
Jan 24 at 12:39
$begingroup$
or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
$endgroup$
– James Arathoon
Jan 24 at 13:20
$begingroup$
First comment should read after partial term 10
$endgroup$
– James Arathoon
Jan 24 at 13:27
$begingroup$
@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
$endgroup$
– Claude Leibovici
Jan 24 at 16:00
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
Considering
$$a_n=frac{H_{2n}}{n^2,4^n}{2n choose n}qquad text{and} qquad S_p=sum_{n=1}^p a_n$$ none of the CAS I tried was able to find an expression for the partial sums or the infinite sum. Numerically, as shown below, the convergence looks to be extremely slow
$$left(
begin{array}{cc}
p & S_p \
1000 & 1.21081501745 \
2000 & 1.21088004598 \
3000 & 1.21089738494 \
4000 & 1.21090493158 \
5000 & 1.21090901996 \
6000 & 1.21091153294 \
7000 & 1.21091321066 \
8000 & 1.21091439815 \
9000 & 1.21091527609 \
10000 & 1.21091594745
end{array}
right)$$ which can be explained by the fact that, for large values of $n$
$$frac {a_{n+1}} {a_n} simeq 1+frac{2-5( log (2n)+ gamma) }{2 n left(log(2n)+gamma right)}$$ For the infinite summation, the result seems to be close to $1.2109201$ which is not identified by inverse symbolic calculators.
For large values of $n$, we also can find
$$a_nsimeq b_n=frac{ log (2n)+gamma }{n^{5/2},sqrt{pi }}$$ which does not help much even if
$$sum_{n=1}^infty b_n=frac{(gamma+log(2)) zeta left(frac{5}{2}right)-zeta 'left(frac{5}{2}right)}{sqrt{pi }}approx 1.18001$$
However, numerically, this can be of some help writing
$$S_infty=S_p+sum_{n=p+1}^infty b_n$$
$$left(
begin{array}{cc}
p & S_infty approx \
100 & 1.2109213325 \
200 & 1.2109203863 \
300 & 1.2109202368 \
400 & 1.2109201900 \
500 & 1.2109201700 \
600 & 1.2109201590 \
700 & 1.2109201535 \
800 & 1.2109201498 \
900 & 1.2109201475 \
1000 & 1.2109201458
end{array}
right)$$
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
$begingroup$
First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
$endgroup$
– James Arathoon
Jan 24 at 12:39
$begingroup$
or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
$endgroup$
– James Arathoon
Jan 24 at 13:20
$begingroup$
First comment should read after partial term 10
$endgroup$
– James Arathoon
Jan 24 at 13:27
$begingroup$
@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
$endgroup$
– Claude Leibovici
Jan 24 at 16:00
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
Considering
$$a_n=frac{H_{2n}}{n^2,4^n}{2n choose n}qquad text{and} qquad S_p=sum_{n=1}^p a_n$$ none of the CAS I tried was able to find an expression for the partial sums or the infinite sum. Numerically, as shown below, the convergence looks to be extremely slow
$$left(
begin{array}{cc}
p & S_p \
1000 & 1.21081501745 \
2000 & 1.21088004598 \
3000 & 1.21089738494 \
4000 & 1.21090493158 \
5000 & 1.21090901996 \
6000 & 1.21091153294 \
7000 & 1.21091321066 \
8000 & 1.21091439815 \
9000 & 1.21091527609 \
10000 & 1.21091594745
end{array}
right)$$ which can be explained by the fact that, for large values of $n$
$$frac {a_{n+1}} {a_n} simeq 1+frac{2-5( log (2n)+ gamma) }{2 n left(log(2n)+gamma right)}$$ For the infinite summation, the result seems to be close to $1.2109201$ which is not identified by inverse symbolic calculators.
For large values of $n$, we also can find
$$a_nsimeq b_n=frac{ log (2n)+gamma }{n^{5/2},sqrt{pi }}$$ which does not help much even if
$$sum_{n=1}^infty b_n=frac{(gamma+log(2)) zeta left(frac{5}{2}right)-zeta 'left(frac{5}{2}right)}{sqrt{pi }}approx 1.18001$$
However, numerically, this can be of some help writing
$$S_infty=S_p+sum_{n=p+1}^infty b_n$$
$$left(
begin{array}{cc}
p & S_infty approx \
100 & 1.2109213325 \
200 & 1.2109203863 \
300 & 1.2109202368 \
400 & 1.2109201900 \
500 & 1.2109201700 \
600 & 1.2109201590 \
700 & 1.2109201535 \
800 & 1.2109201498 \
900 & 1.2109201475 \
1000 & 1.2109201458
end{array}
right)$$
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
This is not an answer but it is too long for a comment.
Considering
$$a_n=frac{H_{2n}}{n^2,4^n}{2n choose n}qquad text{and} qquad S_p=sum_{n=1}^p a_n$$ none of the CAS I tried was able to find an expression for the partial sums or the infinite sum. Numerically, as shown below, the convergence looks to be extremely slow
$$left(
begin{array}{cc}
p & S_p \
1000 & 1.21081501745 \
2000 & 1.21088004598 \
3000 & 1.21089738494 \
4000 & 1.21090493158 \
5000 & 1.21090901996 \
6000 & 1.21091153294 \
7000 & 1.21091321066 \
8000 & 1.21091439815 \
9000 & 1.21091527609 \
10000 & 1.21091594745
end{array}
right)$$ which can be explained by the fact that, for large values of $n$
$$frac {a_{n+1}} {a_n} simeq 1+frac{2-5( log (2n)+ gamma) }{2 n left(log(2n)+gamma right)}$$ For the infinite summation, the result seems to be close to $1.2109201$ which is not identified by inverse symbolic calculators.
For large values of $n$, we also can find
$$a_nsimeq b_n=frac{ log (2n)+gamma }{n^{5/2},sqrt{pi }}$$ which does not help much even if
$$sum_{n=1}^infty b_n=frac{(gamma+log(2)) zeta left(frac{5}{2}right)-zeta 'left(frac{5}{2}right)}{sqrt{pi }}approx 1.18001$$
However, numerically, this can be of some help writing
$$S_infty=S_p+sum_{n=p+1}^infty b_n$$
$$left(
begin{array}{cc}
p & S_infty approx \
100 & 1.2109213325 \
200 & 1.2109203863 \
300 & 1.2109202368 \
400 & 1.2109201900 \
500 & 1.2109201700 \
600 & 1.2109201590 \
700 & 1.2109201535 \
800 & 1.2109201498 \
900 & 1.2109201475 \
1000 & 1.2109201458
end{array}
right)$$
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 24 at 5:40
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
$begingroup$
First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
$endgroup$
– James Arathoon
Jan 24 at 12:39
$begingroup$
or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
$endgroup$
– James Arathoon
Jan 24 at 13:20
$begingroup$
First comment should read after partial term 10
$endgroup$
– James Arathoon
Jan 24 at 13:27
$begingroup$
@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
$endgroup$
– Claude Leibovici
Jan 24 at 16:00
add a comment |
$begingroup$
First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
$endgroup$
– James Arathoon
Jan 24 at 12:39
$begingroup$
or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
$endgroup$
– James Arathoon
Jan 24 at 13:20
$begingroup$
First comment should read after partial term 10
$endgroup$
– James Arathoon
Jan 24 at 13:27
$begingroup$
@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
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– Claude Leibovici
Jan 24 at 16:00
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First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
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– James Arathoon
Jan 24 at 12:39
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First five terms summed of a partial closed form sum approximation achieved using Mathematica 11.3. After term 9 the $log[2]$ term disappears and then appears occasionally after that, not sure if this is an error or feature of this partial series. $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^1 text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n 2^{2 n}}right] , dxright]=frac{11 left(515328 pi ^2-859301-6183936 log ^2(2)right)}{11612160}$$
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– James Arathoon
Jan 24 at 12:39
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or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
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– James Arathoon
Jan 24 at 13:20
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or for slightly simpler integral $$text{FullSimplify}left[2 sum _{m=1}^5 left(H_{2 m}-H_{2 m-2}right) int_0^{frac{1}{2}} text{FullSimplify}left[sum _{n=m}^{infty } frac{binom{2 n}{n} x^{2 n-1}}{n}right] , dxright] $$
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– James Arathoon
Jan 24 at 13:20
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First comment should read after partial term 10
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– James Arathoon
Jan 24 at 13:27
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First comment should read after partial term 10
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– James Arathoon
Jan 24 at 13:27
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@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
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– Claude Leibovici
Jan 24 at 16:00
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@JamesArathoon. Very interesting and I thank you. Would you mind to send me the input form MMA syntax in a .txt file . I should try to play with it when going to university (my e-mail address is my profile). Thanks again :-)
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– Claude Leibovici
Jan 24 at 16:00
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Don't you mean $H_n=sum_{k=1}^{n}frac{1}{k}$?
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– James Arathoon
Jan 24 at 1:13
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Oh yes sorry, now it's corrected !
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– Harmonic Sun
Jan 24 at 1:14
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Would the fact that $frac{{2nchoose n}}{4^n}=frac{1}{2pi}int_0^{2pi} cos^{2n} x,dx$ be useful if you can switch integration and summation?
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– Zachary
Jan 24 at 1:47
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Looks like a very good idea, but as far as I know there is no known closed form for the power series $sumlimits_{n=1}^{infty}frac{H_{2n}}{n^2}x^{2n}$... And even if there were, I guess we would eventually get a very nasty looking integral, especially after composing with the cosine function...
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– Harmonic Sun
Jan 24 at 2:32