TITLE: General formula of $\sum_{n\ge1}{\binom{2n}{n}^{-k}}$.
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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).