# General formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$.

If $k=1$, we have \begin{align} \sum_{n\ge1}\frac{1}{\binom{2n}{n}}&=\frac12\sum_{n\ge1}n\mathrm{B}(n,n)\\ &=\frac12\int _{0}^{1}\sum_{n\ge1}n(t-t^2)^{{n-1}}{\mathrm d}t\\ &=\frac12\int _{0}^{1}\frac{1}{(t-t^2-1)^2}{\mathrm d}t\\ &=\frac13+\frac{2\sqrt3\pi}{27} \end{align} I wonder if there exists a general formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$ with $k\in\Bbb N^*$. This may involve some ${}_pF_q$ functions.

We have $$\binom{2n}{n}^{-1} = \frac{\Gamma(n+1)^2}{\Gamma(2n+1)} = \frac{\Gamma(1/2)\Gamma(n+1)}{4^n\Gamma(n+1/2)} = \frac{(1)_n}{4^n(1/2)_n}$$ by the duplication formula, where $(a)_n = a(a+1)\dotsm(a+n-1)$. Then $$\sum_{n=0}^{\infty} \binom{2n}{n}^{-k} = \sum_{n=0}^{\infty} \frac{((1)_n)^k}{((1/2)_n)^k} (4^{-k})^n = \pi^{k} {}_{k+1}F_k \left(\begin{matrix} 1,1,1,\dotsc,1 \\ 1/2,1/2,\dotsc,1/2 \end{matrix} ; \frac{1}{4^k} \right),$$ by the definition $${}_pF_q \left(\begin{matrix} a_1,a_2,\dotsc,a_p \\ b_1,b_2,\dotsc,b_q \end{matrix} ; z \right) = \sum_{k=0}^{\infty} \frac{(a_1)_n(a_2)_n \dotsm (a_p)_n}{(b_1)_n(b_2)_n \dotsm (b_q)_n} \frac{z^n}{n!}.$$

• Now use this for $k=2$. – marty cohen Apr 19 '18 at 1:11
• @martycohen Indeed! The OP did ask for a general formula, possibly involving hypergeometric functions, not a useful formula... – Chappers Apr 19 '18 at 1:14
• That's why I upvoted your answer. – marty cohen Apr 19 '18 at 2:02
• It'd certainly be nicer if $$\sum_n x^{\binom{2n}{n}}$$ had a closed form we could just integrate a few times. – Chappers Apr 19 '18 at 2:06
• That's a lacunary series, since the exponents grow faster than linearly, so almost certainly not. en.wikipedia.org/wiki/Lacunary_function – marty cohen Apr 19 '18 at 2:39

Chappers already wrote these series as hypergeometric functions, so I will just outline how to deal with the case $k=2$, exhibiting a relation with elliptic integrals. The key identities are $$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx,\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2 \binom{2n}{n}}$$ which have been deeply exploited also here.

The case $k=2$. $$\sum_{n\geq 0}\binom{2n}{n}^{-2}=\phantom{}_2 F_1\left(1,1,1;\tfrac{1}{2},\tfrac{1}{2};\tfrac{1}{16}\right)$$ $$2\arcsin^2\left(\frac{x}{4}\right)=\sum_{n\geq 1}\frac{x^{2n}}{4^n n^2 \binom{2n}{n}},\quad \frac{x\arcsin(x/4)}{\sqrt{1-x^2/16}}=\sum_{n\geq 1}\frac{2 x^{2n}}{4^n n \binom{2n}{n}}$$ $$\frac{x^3}{4-\frac{x^2}{4}}+\frac{x^4\arcsin(x/4)}{16\left(1-x^2/16\right)^{3/2}}+\frac{x^2\arcsin(x/4)}{\sqrt{1-x^2/16}}= \sum_{n\geq 1}\frac{4 x^{2n+1}}{4^n \binom{2n}{n}}$$

$$\frac{x^2 \left(x \sqrt{16-x^2} \left(64-x^2\right)+16 \left(32+x^2\right) \arcsin\left(\frac{x}{4}\right)\right)}{\left(16-x^2\right)^{5/2}}= \sum_{n\geq 1}\frac{(2n+1) x^{2n+1}}{4^n \binom{2n}{n}}$$

$$\sum_{n\geq 1}\binom{2n}{n}^{-2}=\\=\int_{0}^{\pi/2}\frac{\sin^2(\theta) \left(\sin(\theta) \sqrt{16-\sin^2(\theta)} \left(64-\sin^2(\theta)\right)+16 \left(32+\sin^2(\theta)\right) \arcsin\left(\frac{\sin\theta}{4}\right)\right)}{\left(16-\sin^2\theta\right)^{5/2}}\,d\theta\\ = \frac{3}{5}-\frac{32}{5\sqrt{15}}\arcsin\left(\frac{1}{4}\right)+ 16\int_{0}^{1/4}\frac{u^2 \left(2+u^2\right) \arcsin\left(u\right)}{\sqrt{1-16 u^2}\left(1-u^2\right)^{5/2}}\,du$$ where the last integral is related to $E\left(\frac{1}{4}\right)$ and $K\left(\frac{1}{4}\right)$, which can be computed through algorithms with quadratic convergence (Brent-Salamin and the like).