Evaluating the sum of the series $sumlimits_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} $












38












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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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  • $begingroup$
    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
















38












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














38












38








38


10



$begingroup$


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







share|cite|improve this question











$endgroup$




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






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edited Dec 25 '18 at 10:30









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Random VariableRandom Variable

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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


















  • $begingroup$
    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










2 Answers
2






active

oldest

votes


















14












$begingroup$

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
$$
begin{align}
sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
&=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
&=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
&=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
end{align}
$$






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$endgroup$





















    9












    $begingroup$

    $newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
    newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
    newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
    newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
    newcommand{dd}{{rm d}}
    newcommand{ds}[1]{displaystyle{#1}}
    newcommand{expo}[1]{,{rm e}^{#1},}
    newcommand{fermi}{,{rm f}}
    newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
    newcommand{half}{{1 over 2}}
    newcommand{ic}{{rm i}}
    newcommand{iff}{Longleftrightarrow}
    newcommand{imp}{Longrightarrow}
    newcommand{pars}[1]{left(, #1 ,right)}
    newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
    newcommand{pp}{{cal P}}
    newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
    newcommand{sech}{,{rm sech}}
    newcommand{sgn}{,{rm sgn}}
    newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
    newcommand{verts}[1]{leftvert, #1 ,rightvert}$
    $ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$




    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}







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      2 Answers
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      2 Answers
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      active

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      active

      oldest

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      14












      $begingroup$

      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
      $$
      begin{align}
      sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
      &=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
      &=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
      &=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
      end{align}
      $$






      share|cite|improve this answer









      $endgroup$


















        14












        $begingroup$

        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
        $$
        begin{align}
        sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
        &=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
        &=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
        &=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
        end{align}
        $$






        share|cite|improve this answer









        $endgroup$
















          14












          14








          14





          $begingroup$

          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
          $$
          begin{align}
          sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
          &=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
          &=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
          &=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
          end{align}
          $$






          share|cite|improve this answer









          $endgroup$



          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
          $$
          begin{align}
          sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}}
          &=frac{2 pi}{3}operatorname{Cl}_2left(frac{pi}{3}right)+2operatorname{Cl}_3left(frac{pi}{3}right)-2 zeta(3)\
          &=frac{2 pi}{3}frac{sqrt 3}{12}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]+2frac{1}{3}zeta(3)-2zeta(3)\
          &=frac{pisqrt 3}{18}left[psi_1left(frac{1}{3}right)-psi_1left(frac{2}{3}right)right]-frac{4}{3}zeta(3).
          end{align}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 13 '13 at 3:09









          alexjoalexjo

          12.5k1430




          12.5k1430























              9












              $begingroup$

              $newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
              newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
              newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
              newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
              newcommand{dd}{{rm d}}
              newcommand{ds}[1]{displaystyle{#1}}
              newcommand{expo}[1]{,{rm e}^{#1},}
              newcommand{fermi}{,{rm f}}
              newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
              newcommand{half}{{1 over 2}}
              newcommand{ic}{{rm i}}
              newcommand{iff}{Longleftrightarrow}
              newcommand{imp}{Longrightarrow}
              newcommand{pars}[1]{left(, #1 ,right)}
              newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
              newcommand{pp}{{cal P}}
              newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
              newcommand{sech}{,{rm sech}}
              newcommand{sgn}{,{rm sgn}}
              newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
              newcommand{verts}[1]{leftvert, #1 ,rightvert}$
              $ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$




              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}







              share|cite|improve this answer











              $endgroup$


















                9












                $begingroup$

                $newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
                newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
                newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
                newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
                newcommand{dd}{{rm d}}
                newcommand{ds}[1]{displaystyle{#1}}
                newcommand{expo}[1]{,{rm e}^{#1},}
                newcommand{fermi}{,{rm f}}
                newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
                newcommand{half}{{1 over 2}}
                newcommand{ic}{{rm i}}
                newcommand{iff}{Longleftrightarrow}
                newcommand{imp}{Longrightarrow}
                newcommand{pars}[1]{left(, #1 ,right)}
                newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                newcommand{pp}{{cal P}}
                newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
                newcommand{sech}{,{rm sech}}
                newcommand{sgn}{,{rm sgn}}
                newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
                newcommand{verts}[1]{leftvert, #1 ,rightvert}$
                $ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$




                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}







                share|cite|improve this answer











                $endgroup$
















                  9












                  9








                  9





                  $begingroup$

                  $newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
                  newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
                  newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
                  newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
                  newcommand{dd}{{rm d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,{rm e}^{#1},}
                  newcommand{fermi}{,{rm f}}
                  newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
                  newcommand{half}{{1 over 2}}
                  newcommand{ic}{{rm i}}
                  newcommand{iff}{Longleftrightarrow}
                  newcommand{imp}{Longrightarrow}
                  newcommand{pars}[1]{left(, #1 ,right)}
                  newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{pp}{{cal P}}
                  newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
                  newcommand{sech}{,{rm sech}}
                  newcommand{sgn}{,{rm sgn}}
                  newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert, #1 ,rightvert}$
                  $ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$




                  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}







                  share|cite|improve this answer











                  $endgroup$



                  $newcommand{angles}[1]{leftlangle, #1 ,rightrangle}
                  newcommand{braces}[1]{leftlbrace, #1 ,rightrbrace}
                  newcommand{bracks}[1]{leftlbrack, #1 ,rightrbrack}
                  newcommand{ceil}[1]{,leftlceil, #1 ,rightrceil,}
                  newcommand{dd}{{rm d}}
                  newcommand{ds}[1]{displaystyle{#1}}
                  newcommand{expo}[1]{,{rm e}^{#1},}
                  newcommand{fermi}{,{rm f}}
                  newcommand{floor}[1]{,leftlfloor #1 rightrfloor,}
                  newcommand{half}{{1 over 2}}
                  newcommand{ic}{{rm i}}
                  newcommand{iff}{Longleftrightarrow}
                  newcommand{imp}{Longrightarrow}
                  newcommand{pars}[1]{left(, #1 ,right)}
                  newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
                  newcommand{pp}{{cal P}}
                  newcommand{root}[2]{,sqrt[#1]{vphantom{large A},#2,},}
                  newcommand{sech}{,{rm sech}}
                  newcommand{sgn}{,{rm sgn}}
                  newcommand{totald}[3]{frac{{rm d}^{#1} #2}{{rm d} #3^{#1}}}
                  newcommand{verts}[1]{leftvert, #1 ,rightvert}$
                  $ds{sum_{n=1}^{infty} frac{1}{n^{3} binom{2n}{n}} = frac{ pi sqrt{3}}{18} Big[ psi_{1} left(frac{1}{3} right) - psi_{1} left(frac{2}{3} right) Big]- frac{4}{3} zeta(3)}$




                  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}








                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 27 '18 at 16:55

























                  answered Sep 17 '14 at 4:17









                  Felix MarinFelix Marin

                  69.3k7110147




                  69.3k7110147






























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