TITLE: Exercise from Comtet's Advanced Combinatorics: prove $27\sum_{n=1}^{\infty }1/\binom{2n}{n}=9+2\pi \sqrt{3}$
QUESTION [24 upvotes]: In exercise 36 Miscellaneous Taylor Coefficients using Bernoulli
numbers on pages 88-89 of Louis Comtet's Advanced Combinatorics, 1974,
one is asked to obtain the following explicit formula for the Bernoulli numbers:
$$B_{2n}=(-1)^{n-1}\dfrac{1+\left[ \varphi _{n}\right] }{2(2^{2n}-1)},$$
where
$$\varphi _{n}=\dfrac{2(2^{2n}-1)(2n)!}{2^{2n-1}\pi ^{2n}}\displaystyle\sum_{k=1}^{3n}\dfrac{1}{k^{2n}}$$
(with $\displaystyle\sum_{n\geq 0}B_{n}\dfrac{t^{n}}{n!}=\dfrac{t}{e^{t}-1}$), and to prove, among other sums, that
$$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{\dbinom{2n}{n}}=\dfrac{1}{3}+\dfrac{2\pi\sqrt{3}}{27}.\qquad (\ast )$$
Alfred van der Poorten wrote here (section 10): seeing that
$$\displaystyle\sum_{n=1}^{\infty}\dfrac{x^{2n}}{n^{2}\dbinom{2n}{n}}=2\arcsin^{2}\left( \dfrac{x}{2}\right) \qquad (\ast \ast )$$
(...) formula [ $(\ast )$ ] become[s] quite accessible to proof."
I am not able to show formula $(\ast \ast )$ neither how it can be used to
prove $(\ast )$.

Question: Could you provide (a) more detailed hint(s) on how and/or
different ways in which formula $(\ast )$ can be derived?

Added: For information the other sums are:
$$\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n\dbinom{2n}{n}}=\dfrac{\pi \sqrt{3}}{9},\quad\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{2}\dbinom{2n}{n}}=\dfrac{\pi ^{2}}{18},\quad\displaystyle\sum_{n=2}^{\infty }\dfrac{1}{n^{4}\dbinom{2n}{n}}=\dfrac{17\pi ^{4}}{3240}.$$

REPLY [8 votes]: I'm surprised no one has yet posted the generating function approach.  An excellent reference is Sprugnoli, "Sums of Reciprocals of the Central Binomial Coefficients," Integers, Article A27, 2006.
He proves that the generating function of $4^n \binom{2n}{n}^{-1}$ is
$$Z(t) = \frac{1}{1-t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}} + \frac{1}{1-t}.$$
Substituting $t=1/4$ immediately yields $$\sum_{n=0}^{\infty} \binom{2n}{n}^{-1} = \dfrac{4}{3}+\dfrac{2\pi\sqrt{3}}{27}.$$
He also derives generating functions for $\frac{4^n}{n} \binom{2n}{n}^{-1}$, $\frac{4^n}{n^2} \binom{2n}{n}^{-1}$, and several other similar expressions involving the reciprocals of the central binomial coefficients, which allows him to deduce a few dozen expressions for various finite and infinite sums involving $\binom{2n}{n}^{-1}$.  Again, it's an excellent reference.
Incidentally, in the introduction the author states that his motivation for writing the paper was the very exercise in Comtet that motivated this question.