# Evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$. [duplicate]

Evaluate $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}.$$

My work so far and background to the problem.

This question was inspired by the second page of this paper. The author of the paper doesn't mention how he managed to prove the $$4$$ identities (shown below) at the top of the second page of the paper, so I tried to find my own method of doing so.

\begin{align} \sum\limits_{n=1}^{\infty}\binom{2n}{n}^{-1} &= \frac{1}{3}+\frac{2\pi\sqrt3}{27} \\ \sum\limits_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}^{-1} &= \frac{\pi\sqrt3}{9} \\ \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\binom{2n}{n}^{-1} &= \frac{\pi^2}{18} \\ \sum\limits_{n=1}^{\infty}\frac{1}{n^4}\binom{2n}{n}^{-1} &= \frac{17\pi^4}{3240} \end{align} The method I used was similar to the method the author used in the rest of his paper: trying to find the generating functions for particular sequences. Using this method, like the author I was successful in proving the first $$3$$ identities, since I obtained the following: \begin{align}\sum\limits_{n=1}^{\infty}\binom{2n}{n}^{-1}x^n&=4\sqrt\frac{x}{(4-x)^3}\arcsin\frac{\sqrt{x}}{2}+\frac{x}{4-x}\\ \sum\limits_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}^{-1}x^n&=2\sqrt{\frac{x}{4-x}}\arcsin\frac{\sqrt x}{2}\\ \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\binom{2n}{n}^{-1}x^n&=2\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\end{align} I obtained the second and third power series by dividing the power series above each one by $$x$$ and then integrating.

I then attempted to find a closed form for $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}x^n$$ in order to evaluate $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$$. The way I tried to do this was finding $$\int\frac{2}{x}\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\mathrm{d}x$$ but according to WA this has no solution in terms of elementary functions and it involves some polylogarithms which I am not familiar with at all really, especially when they have a complex argument.

I then realized that $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}=\int_0^1\frac{2}{x}\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\mathrm{d}x=\int_0^{1/2}\frac{4}{u}\left(\arcsin u\right)^2\mathrm{d}u$$ (where the second equality is found by using the substitution $$u=\frac{\sqrt{x}}{2}$$) which could be more helpful since definite integrals can often be evaluated in terms of elementary functions even if the indefinite integral cannot.

From there however I could not think of a way of continuing.

• WolframAlpha gives a closed-form solution HERE in terms of the Riemann Zeta function and the first derivative of the Digamma Function. May 19, 2021 at 18:24
• I'm not sure if any nice closed form exists. You can compute this integral numerically (using sage, say), and get 0.522946192133335. If we plug this into an inverse symbolic calculator the only symbolic representation it knows about is exactly the sum you're trying to compute. May 19, 2021 at 18:26
• One avenue you might try is using hypergeometric identities to massage your sum into a closed form, since we know that it's hypergeometric. Indeed, sage again tells us that your sum is the following hypergeometric function: $$\frac{1}{2} \, \,_4F_3\left(\begin{matrix} 1,1,1,1 \\ 2,2,\frac{3}{2} \end{matrix} ; \frac{1}{4} \right).$$ I know there are a lot of known identities for hypergeometric functions, but I don't have any references since I've never personally needed them. May 19, 2021 at 18:27
• Mathematica/Sage results suggest there is a good reason why $n^3$ is skipped in your referenced paper.
– Momo
May 19, 2021 at 18:34
• math.stackexchange.com/q/499926 Jun 18, 2021 at 14:32

Here is an explicit way to proceed, we are calculating the last integral in the OP, denoted here by $$J$$: \begin{aligned} J &= \int_0^{1/2}\frac 4u\;\arcsin^2 u\; du\qquad\text{(Substitution: u=\sin x)}\\ &= \int_0^a\frac 4{\sin x}\;x^2\; \cos x\; dx\qquad\text{(where a=\arcsin (1/2)=\pi/6)}\\ &= 4 \int_0^a\frac x{\tan x}\;x\; dx\\ &= 4 \int_0^a\left(\frac {2ix}{e^{2ix}-1}+ix\right)\;x\; dx\\ &= 4\Re \int_0^a\frac {2ix}{e^{2ix}-1}\;x\; dx\\ &= -\Re \int_0^a\frac {2ix}{e^{2ix}-1}\;2ix\; d(2ix)\\ &= -\Re \int_0^{2ia}\frac {X^2}{e^X-1}\;dX\\ &= -\Re \left[ \ \color{gray}{ -\frac 13X^3} + X^2\log(1-e^X) + 2X\operatorname{Li}_2(e^X) - 2\operatorname{Li}_3(e^X) \ \right]_0^{2ia} \\ &= -\Re \left[ \ -a^2\log(1-e^{2ia}) + 4ia\operatorname{Li}_2(e^{2ia}) - 2\operatorname{Li}_3(e^{2ia}) + 2\operatorname{Li}_3(e^0) \ \right] \\ &= -\Re \left[ \ 0 + 4ia\operatorname{Li}_2(e^{2ia}) - 2\operatorname{Li}_3(e^{2ia}) + 2\zeta(3) \ \right] \\ &= -\Re \left[ \ 4ia\operatorname{Li}_2(b) - 2\operatorname{Li}_3(b) + 2\zeta(3) \ \right] \ . \end{aligned} The logarithmic term disappeared because $$\Re \log(1-e^{2ia}) = \log |1-e^{2ia}| = \log |1-e^{2i\pi/6}| = \log |e^{-2i\pi/6}| =\log 1 =0\ .$$ The value of $$b$$ is explicitly $$b = e^{2ia} =e^{2i\pi/6} =e^{i\pi/3} \ ,$$ so $$b$$ is the above primitive root of unity of order six.

One can show the relation: $$\Re \operatorname{Li}_3(b) =\frac 13\operatorname{Li}_3(1) =\frac 13\zeta(3)\ .$$ This is seen for instance after a series expansion: \begin{aligned} &2\Re\operatorname{Li}_3(b) = \operatorname{Li}_3(b) + \operatorname{Li}_3(\bar b) \\ &= \frac 1{1^3}\underbrace{(b+\bar b)}_{=1} + \frac 1{2^3}\underbrace{(b^2+\bar b^2)}_{=-1} + \frac 1{3^3}\underbrace{(b^3+\bar b^3)}_{=-2} + \frac 1{4^3}\underbrace{(b^4+\bar b^4)}_{=-1} + \frac 1{5^3}\underbrace{(b^5+\bar b^5)}_{=1} + \frac 1{6^3}\underbrace{(b^6+\bar b^6)}_{=2} + \dots\text{ with 6-periodic repetitions} \\ &= \frac 1{1^3}(1) + \frac 1{2^3}(1-\color{blue}{2}) + \frac 1{3^3}(1-\color{darkgreen}{3}) + \frac 1{4^3}(1-\color{blue}{2}) + \frac 1{5^3}(1) + \frac 1{6^3}(1-\color{blue}{2}-\color{darkgreen}{3}+\color{red}{6}) + \dots\text{ with 6-periodic repetitions} \\ &= \zeta(3)\left( 1 - \frac{\color{blue}{2}}{2^3} - \frac{\color{darkgreen}{3}}{3^3} + \frac{\color{red}{6}}{6^3} \right) \\ &= \zeta(3) \left( 1 - \frac{\color{blue}{2}}{2^3} \right) \left( 1- \frac1{\color{darkgreen}{3}}{3^3} \right) =\zeta(3)\cdot\frac 34\cdot\frac 89 =\zeta(3)\cdot\frac 23\ . \end{aligned} In a similar manner we can (try to) look into the dilogarithmic part. \begin{aligned} &2\Re i\operatorname{Li}_2(b) =\Re i( \operatorname{Li}_2(b) - \operatorname{Li}_2(\bar b)) \\ &= \frac i{1^2}\underbrace{(b-\bar b)}_{=i\sqrt 3} + \frac i{2^2}\underbrace{(b^2-\bar b^2)}_{=i\sqrt 3} + \frac i{3^2}\underbrace{(b^3-\bar b^3)}_{=0} + \frac i{4^2}\underbrace{(b^4-\bar b^4)}_{=-i\sqrt 3} + \frac i{5^2}\underbrace{(b^5-\bar b^5)}_{=-i\sqrt 3} + \frac i{6^2}\underbrace{(b^6-\bar b^6)}_{=0} + \dots\text{ with 6-periodic repetitions} \\ &= \sqrt3\left( \frac1{1^2} + \frac 1{2^2} - \frac1{4^2} - \frac 1{5^2} + \dots \right)\text{ with 6-periodic repetitions} \\ &= \frac {\sqrt3}{18} \left( \psi'\left(\frac 16\right) + \psi'\left(\frac 13\right) \right) -\frac {4 {\sqrt3}}{27}\pi^2 \ . \end{aligned} We have a thus some formula for the expression involving corresponding special values $$L(2,\chi)$$ of the Dirichlet $$L$$-function computed in $$2$$ w.r.t. some characters $$\chi$$ with periodicity modulo six, and in general we can expect some $$\psi$$-values as in:

https://mathworld.wolfram.com/DirichletL-Series.html

We can do slightly better. Using $$b^4=-b$$ and the dilogarithmic relation $$\operatorname{Li}_2(x)+\operatorname{Li}_2(-x)=\frac 12\operatorname{Li}_2(x^2)$$, we obtain: $$\operatorname{Li}_2(b)+\operatorname{Li}_2(b^4)=\frac 12\operatorname{Li}_2(b^2)$$. Here, $$b^2$$ and $$b^4$$ are cubic primitive units, so the periodicity is reduced. Writing series for instance, we get $$\operatorname{Li}_2(b^4)= -\operatorname{Li}_2(b^2)$$. This gives: \begin{aligned} \Im \operatorname{Li}_2(b) &=\frac 32 \Im \operatorname{Li}_2(b^2) \\ &= \frac {3\sqrt 3}4 \left( \frac1{1^2} - \frac 1{2^2} + \frac1{4^2} - \frac 1{5^2} + \dots \right)\text{ with 3-periodic repetitions} \\ &= \frac {3\sqrt 3}4 \left( \frac 29\psi'\left(\frac 13\right) -\frac4{27}\pi^2 \right)\ . \end{aligned} Here, for instance, the sum of the inverses of $$1^2$$, $$4^2$$, $$7^2$$, ... is $$\frac 19\psi'\left(\frac 13 \right)$$.

Putting all together: \color{blue}{ \begin{aligned} J &= \frac 23\pi\cdot \Im \operatorname{Li}_2(b) - \frac 43\zeta(3) \\ &= \pi\cdot \Im \operatorname{Li}_2(b^2) - \frac 43\zeta(3) \ . \end{aligned} } (And the dilogarithmic values have expressions in terms of $$\psi'$$ computed in $$1/3$$.)

Numerical check using pari/gp:

? \p 50
? J = intnum( u=0, 1/2, 4/u*asin(u)^2);
? b = exp(I*Pi/3);
? J
%158 = 0.52294619213333510849118518352730354016304459174398
? 2/3*Pi*imag(dilog(b)) - 4/3*zeta(3)
%159 = 0.52294619213333510849118518352730354016304459174398
?     Pi*imag(dilog(b^2)) - 4/3*zeta(3)
%160 = 0.52294619213333510849118518352730354016304459174398


Numerical check using sage:

sage: var('x,u,k');
sage: J = integral( 4/u * arcsin(u)^2, u, 0, 1/2, hold=True )
sage: J.n()
0.5229461921333352
sage: b = (1 + i*sqrt(3))/2
sage: ( 2/3 * pi * imag(dilog(b)) - 4/3 * zeta(3) ).n( digits=50 )
0.52294619213333510849118518352730354016304459174398
sage: (       pi * imag(dilog(b^2)) - 4/3 * zeta(3) ).n( digits=50 )
0.52294619213333510849118518352730354016304459174398

sage: imag( dilog(b) ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303

sage: imag( 3/2 * dilog(b^2) ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303

sage: ( sum( 1/(3*k+1)^2 - 1/(3*k+2)^2, k, 0, oo) * 3*sqrt(3)/4 ).n( digits=50 )
1.0149416064096536250212025542745202859416893075303

sage:   sum( 1/(3*k+1)^2 - 1/(3*k+2)^2, k, 0, oo)
-4/27*pi^2 + 2/9*psi(1, 1/3)

sage: var('X');
sage: integral( X^2 / (exp(X) - 1), X)
-1/3*X^3 + X^2*log(-e^X + 1) + 2*X*dilog(e^X) - 2*polylog(3, e^X)


(Please omit this coding section, if this feels misplaced, it was done only for my calm sleep, so that i can easily double check the computations next day with better eyes.)

Hint: We find in Evaluations of Binomial Series by J. M. Borwein and R. Girgensohn the formula (47) \begin{align*} \sum_{n\geq 1}\frac{1}{n^3\binom{2n}{n}}=\frac{2}{3}\pi\Im\left(L_2\left(e^{i\pi/3}\right)\right) -\frac{4}{3}\zeta(3) \end{align*} with $$L_p(z)=\sum_{n>0}\frac{z^n}{n^p}$$ the polylogarithms.