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.
 A: 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
$$ {}_pF_q \left(\begin{matrix} a_1,a_2,\dotsc,a_p \\ b_1,b_2,\dotsc,b_q \end{matrix} ; z \right) = \sum_{k=0}^{\infty} \frac{(a_1)_n(a_2)_n \dotsm (a_p)_n}{(b_1)_n(b_2)_n \dotsm (b_q)_n} \frac{z^n}{n!}. $$
A: 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).
