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Sums which involve the central binomial coefficient $\binom{2n}{n} = \frac{(2n)!}{(n!)^2}$ and its inverse are actually quite common and often have nice closed forms. For example, $$\begin{eqnarray} \sum _{n=1} ^{\infty} \frac{4^n}{n^2\binom{2n}{n}} &=& \frac{\pi^2}{2}\\ \sum _{n=1} ^{\infty} \frac{(-1)^n}{n^3\binom{2n}{n}} &=& -\frac{2}{5} \zeta (3), \quad \zeta(3) \text{ is Apery's constant}\\ \sum _{n=0 } ^{\infty} \frac{4^n}{(2n+1)^2\binom{2n}{n}} &=& 2\rm G, \quad \rm G \text{ denotes Catalan's constant}\\ \end{eqnarray}$$ Also see this question and that question for more identities of this form. Now, my question is, what is the closed form for the sum $$\sum _{n=0 } ^{\infty} \frac{4^n}{(2n-1)^2\binom{2n}{n}} \approx 3.95462$$ By closed form I hope to mean an expression without hypergeometric functions. Sometimes these series could be represented as integrals involving logarithms, for this reason see also this hard to read but wonderful article on logarithmic integrals and related sums.

Nikitan
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  • Wolfram gives $$\sum_{n=0}^{\infty} \frac{4^n}{(2n-1)^2 \binom{2n}{n}} = {}_4F_3\left(\frac{1}{2}, \frac{1}{2}, 1, 1; \frac{3}{2}, \frac{3}{2}, 2; 1 \right) \approx 3.90475 $$ – Leox Aug 20 '24 at 15:28
  • @Leox, as I've said in the post I am looking for a non hypergeometric closed form, the worst function I would welcome is the polylogarithm, and even he would spoil the party somewhat – Nikitan Aug 20 '24 at 15:56
  • And also, I get $\mbox{}{4}!\operatorname{F}_3\left(1,1,\frac{1}{2},\frac{1}{2};\ \frac{3}{2},\frac{3}{2},2;\ 1\right) = \frac{1}{4}(\pi^2 -4 \log (4)) \approx 1.08111$. Maybe you've meant $\mbox{}{4}!\operatorname{F}_3\left(1,1,-\frac{1}{2},-\frac{1}{2};\ \frac{1}{2},\frac{1}{2},\frac{1}{2};\ 1\right)$? – Nikitan Aug 20 '24 at 15:59
  • Sorry: $$ \sum_{n=0}^{\infty} \frac{4^n}{(2n-1)^2 \binom{2n}{n}} = {}_4F_3\left(-\frac{1}{2}, -\frac{1}{2}, 1, 1; \frac{1}{2}, \frac{1}{2}, \frac{1}{2}; 1 \right) \approx 3.954621213 $$ – Leox Aug 20 '24 at 16:53
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    Okay! I actually managed to solve it: $$S = 1 + 2 \rm G + \frac{3}{16}\pi^3 + \frac{1}{4}\pi (\log 2)^2 - 4 \Im \operatorname{Li}_3 (1 + i) \approx 3.95462$$ – Nikitan Aug 21 '24 at 11:57
  • write your solution as own answer – Leox Aug 21 '24 at 15:11

1 Answers1

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Consider the summand $$X_n = \frac{4^n}{(2n-1)^2} \binom{2n}{n}^{-1}$$ Then notice that $$X_{n+1} = \frac{4^{n+1}}{(2n+1)^2}\binom{2n+2}{n+1}^{-1} = \frac{2n+2}{(2n+1)^3} \cdot 4^n\binom{2n}{n}^{-1}$$ Now, by partial fraction decomposition $$X_{n+1} = \frac{4^n}{(2n+1)^2}\binom{2n}{n}^{-1} + \frac{4^n}{(2n+1)^3}\binom{2n}{n}^{-1}$$ Therefore the sum $$\sum _{n=0} ^{\infty} X_n = \sum _{n=-1} ^{\infty} X_{n+1} = 1 + \sum _{n=0} ^{\infty} X_{n+1} = \\ = 1+ \sum _{n=0} ^{\infty} \frac{4^n}{(2n+1)^2} \binom{2n}{n}^{-1} + \sum _{n=0} ^{\infty} \frac{4^n}{(2n+1)^3} \binom{2n}{n}^{-1}= \\ =1+2\rm G + \frac{3}{16}\pi ^3 + \frac{1}{4} \pi (\log 2)^2 - 4\Im \operatorname{Li}_3(1+i) $$ where $\operatorname{Li}_n(x)$ is the polylogarithm. The last sum is taken from my amateur-ish answer to this question.

Nikitan
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