Evaluating the sum of the series $sumlimits_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} $
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Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
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Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
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Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
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– Gerry Myerson
Sep 21 '13 at 1:15
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That sum is the alternating version of this one.
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– Random Variable
Sep 21 '13 at 1:31
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And the techniques that apply to the alternating version don't work for the one you ask about?
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– Gerry Myerson
Sep 21 '13 at 1:41
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They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
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– Random Variable
Sep 21 '13 at 1:56
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use Clausen function
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– alexjo
Nov 13 '13 at 21:08
add a comment |
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Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
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Wolfram MathWorld states that $$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Bigg[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg]- frac{4}{3} zeta(3) , , $$
where $psi_{1}(x)$ is the trigamma function.
I can't seem to get the answer in that form.
Using the Taylor expansion $ displaystyle arcsin^{2}(x) = frac{1}{2} sum_{n=1}^{infty} frac{1}{n^{2} binom{2n}{n}} (2x)^{2n}$, I get
$$ sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx. $$
Then I integrating by parts, I get
$$ begin{align} &4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} , dx \ &= frac{pi^{2}}{9} ln left(frac{1}{2} right) - 8 int_{0}^{frac{1}{2}} frac{arcsin (x) ln (x)}{sqrt{1-x^{2}}} , dx \ &= - frac{pi^{2}}{9} ln 2 - 8 int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 ln 2 int_{0}^{frac{pi}{6}} u , du - 8int_{0}^{frac{pi}{6}} u ln (sin u ) , du \ &= - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) , du \ &= -8 text{Re} int_{0}^{frac{pi}{6}} u ln (1-e^{2iu}) , du \ &= 8 text{Re} int_{0}^{frac{pi}{6}} u sum_{n=1}^{infty} frac{e^{2in u}}{n} , du \ &= 8 sum_{n=1}^{infty} frac{1}{n} int_{0}^{frac{pi}{6}} u cos (2nu) , du \ &= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3) \ &= frac{2 pi}{3} Bigg( frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{2}} + frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{2}} - frac{sqrt{3}}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{2}} Bigg) \ &+ 2 Bigg( frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+1)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+2)^{3}} - sum_{n=0}^{infty} frac{1}{(6n+3)^{3}} - frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+4)^{3}} \ &+ frac{1}{2} sum_{n=0}^{infty} frac{1}{(6n+5)^{3}} + sum_{n=1}^{infty} frac{1}{(6n)^{3}} Bigg) - 2 zeta(3) \ &= frac{pi sqrt{3}}{108} Bigg( psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) Bigg) + frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &-28 zeta(3) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) + 4 zeta(3)Bigg) - 2 zeta (3) .end{align}$$
From here I've been going in circles trying to get the answer in the form given above.
EDIT:
Using the duplication formula for the trigamma function (i.e., $ displaystyle 4 psi_{1}(2x) = psi_{1}(x) + psi_{1} left(x + frac{1}{2} right) $),
$$ begin{align} &psi_{1} left(frac{1}{6} right) + psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) - psi_{1}left(frac{5}{6} right) \ &= 4 psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) -psi_{1} left(frac{2}{3} right) - 4 psi_{1} left(frac{2}{3} right) + psi_{1} left(frac{1}{3} right) \ &= 6 psi_{1} left(frac{1}{3} right) - 6 psi_{1} left(frac{2}{3} right). end{align}$$
Therefore,
$$ begin{align} sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} &= frac{sqrt{3} pi}{18} Bigg( psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Bigg) +
frac{1}{432} Bigg( - psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) \ &+ psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) - 2 zeta(3)Bigg) - 2 zeta (3). end{align}$$
So it comes down to somehow showing that $$-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) = 312 zeta(3) .$$
SECOND EDIT:
Using the duplication formula for $psi_{2}(x)$, we get
$$ begin{align} &-psi_{2} left(frac{1}{6} right) + psi_{2} left(frac{1}{3} right) + psi_{2} left(frac{2}{3} right) - psi_{2} left(frac{5}{6} right) \ &= -8 psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) +psi_{2} left(frac{2}{3} right) - 8 psi_{2} left(frac{2}{3} right) + psi_{2} left(frac{1}{3} right) \ &= -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) . end{align}$$
And using the more general multiplication formula, we get
$$ psi_{2} left(frac{1}{3} right) + psi_{2} left( frac{2}{3} right) + psi_{2}(1) = 27 psi_{2} (1) .$$
Therefore,
$$ -6 psi_{2} left(frac{1}{3} right) - 6 psi_{2} left(frac{2}{3} right) = -156 psi_{2} (1) = 312 zeta(3) .$$
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
sequences-and-series complex-analysis special-functions binomial-coefficients gamma-function
edited Dec 25 '18 at 10:30
Did
249k23229468
249k23229468
asked Sep 20 '13 at 21:20
Random VariableRandom Variable
25.6k173139
25.6k173139
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
add a comment |
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
newcommand{dd}{{rm d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,{rm e}^{#1},}
newcommand{fermi}{,{rm f}}
newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
newcommand{half}{{1 over 2}}
newcommand{ic}{{rm i}}
newcommand{iff}{Longleftrightarrow}
newcommand{imp}{Longrightarrow}
newcommand{pars}[1]{left(, #1 ,right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{pp}{{cal P}}
newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
newcommand{sech}{,{rm sech}}
newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
$endgroup$
add a comment |
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
$endgroup$
add a comment |
$begingroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
$endgroup$
Using the Taylor expansion for $arcsin^2(x)$ and integrating, it's easy to show (as you made) that
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&= 4 int_{0}^{frac{1}{2}} frac{arcsin^{2}(x)}{x} operatorname{d}x = - 8 int_{0}^{frac{pi}{6}} u ln ( 2 sin u ) operatorname{d}u \
&= frac{2 pi}{3} sum_{n=1}^{infty} frac{sin (frac{n pi}{3})}{n^{2}} + 2 sum_{n=1}^{infty} frac{cos (frac{n pi}{3})}{n^{3}} - 2 zeta(3)\
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)
end{align}
$$
recalling that the Clausen function are defined as
$$
operatorname{Cl}_{m}(theta)=
begin{cases}displaystyle
sum_{n=1}^{infty} frac{sin (ntheta)}{n^{m}}& text{for } mtext{ even}\
displaystylesum_{n=1}^{infty} frac{cos(ntheta)}{n^{m}}& text{for } mtext{ odd}
end{cases}
$$
The value of the Clausen function $operatorname{Cl}_3$ at $frac{pi}{3}$ is
$$
operatorname{Cl}_3left(frac{pi}{3}right)=frac{1}{2}left(1-2^{-2}right)left(1-3^{-2}right)zeta(3)=frac{1}{3}zeta(3)
$$
From the duplication formula
$$
operatorname{Cl}_{2m}(2theta)=2^{2m-1}left[operatorname{Cl}_{2m}(theta)-operatorname{Cl}_{2m}(pi-theta)right]
$$
we find
$$
operatorname{Cl}_{2}left(frac{2pi}{3}right)=frac{2}{3}operatorname{Cl}_{2}left(frac{pi}{3}right).
$$
From the identities for the trigamma function at $1/3$ and $2/3$
$$
begin{align}
psi_1left(frac{1}{3}right) &=frac{2pi^2}{3}+3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)\
psi_1left(frac{2}{3}right) &=frac{2pi^2}{3}-3sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
end{align}
$$
one has
$$
psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)=6sqrt 3operatorname{Cl}_{2}left(frac{2pi}{3}right)
$$
and then
$$
operatorname{Cl}_{2}left(frac{pi}{3}right)=frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]
$$
Finally, putting all together, we have
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$
answered Nov 13 '13 at 3:09
alexjoalexjo
12.5k1430
12.5k1430
add a comment |
add a comment |
$begingroup$
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$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
$endgroup$
add a comment |
$begingroup$
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$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
$endgroup$
add a comment |
$begingroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
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$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
$endgroup$
$newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
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newcommand{pars}[1]{left(, #1 ,right)}
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newcommand{sgn}{,{rm sgn}}
newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
newcommand{verts}[1]{leftvert, #1 ,rightvert}$
$ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n{2n choose n}}
=sum_{n = 1}^{infty}x^{n},
{Gammapars{n}Gammapars{n + 1} over Gammapars{2n + 1}}
=sum_{n = 1}^{infty}x^{n}int_{0}^{1}t^{n - 1}pars{1 - t}^{n},dd t
\[3mm] & =
int_{0}^{1}sum_{n = 1}^{infty}bracks{txpars{1 - t}}^{n}
,{dd t over t}
=int_{0}^{1}bracks{{1 over 1 - txpars{1 - t}} - 1},{dd t over t}
\[3mm] & =
int_{0}^{1}{xpars{1 - t} over 1 - txpars{1 - t}},dd t
end{align}
begin{align}
&sum_{n = 1}^{infty}{x^{n} over n^{2}{2n choose n}}
=int_{0}^{x}dd yint_{0}^{1}{1 - t over 1 - typars{1 - t}},dd t
=-int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
end{align}
begin{align}
&color{#66f}{largesum_{n = 1}^{infty}{1 over n^{3}{2n choose n}}}
=-int_{0}^{1}{dd x over x}
int_{0}^{1}{lnpars{1 - bracks{1 - t}tx} over t},dd t
\[3mm] = &
int_{0}^{1}
{{rm Li}_{2}pars{tbracks{1 - t}} over t},dd t
\[3mm] = &
color{#66f}{{pi over 36root{3}}, left[psi ^{(1)}left(frac{1}{3}right)-psi ^{(1)}left(frac{2}{3}right)+psi ^{(1)}left(frac{1}{6}right)-psi ^{(1)}left(frac{5}{6}right)right]-frac{4 zeta (3)}{3}}
\[3mm] approx & {tt 0.5229}
end{align}
edited Dec 27 '18 at 16:55
answered Sep 17 '14 at 4:17
Felix MarinFelix Marin
69.3k7110147
69.3k7110147
add a comment |
add a comment |
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$begingroup$
Isn't this the sum that came up in Apery's proof of the irrationality of $zeta(3)$? If so, it may be in Alf van der Poorten's paper, "A Proof that Euler Missed". Also, I think the sum may be done in Comtet's book on Combinatorics. Also, look at the "Related" questions running down the side of this page.
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:15
$begingroup$
That sum is the alternating version of this one.
$endgroup$
– Random Variable
Sep 21 '13 at 1:31
$begingroup$
And the techniques that apply to the alternating version don't work for the one you ask about?
$endgroup$
– Gerry Myerson
Sep 21 '13 at 1:41
$begingroup$
They work to a point. But the integrand of the other one is $ displaystyle frac{text{arcsinh}^{2}(x)}{x}$.
$endgroup$
– Random Variable
Sep 21 '13 at 1:56
$begingroup$
use Clausen function
$endgroup$
– alexjo
Nov 13 '13 at 21:08