Calculate $\sum_{n = 0}^\infty \frac{C_n^2}{16^n}$

Question

My question is how to prove that $$ \sum_{n = 0}^\infty \frac{C_n^2}{16^n} = \frac{16}{\pi}-4 $$ where $$ C_n = \frac{1}{n+1} \binom{2n}{n}, \quad n=0,1,2,\dots $$ is the $n$-th Catalan number.

Motivation. This question shows up when I played with the infinte series $f(z)=\sum_{n=0} C_n^2 z^n$, which has radius of convergence $1/16$. This series does not seem to have a nice general formula, in contrast to $\sum_{n=0}^\infty C_n z^n$ (Wolfram gives a complicated expression involving Gamma and hypergeometric functions). However, at $z=1/16$, the series $f(z)$ obtains this simple value.

Answer 1

This avoids Parseval's Theorem. Prove by induction or observing that the partial sum telescopes:

$$\sum_{n=0}^{k}{\binom{2n}{n}^{2}\over 16^{n}(n+1)^2}=-4+\frac{(5+4k)\binom{2k+1}{k}^2}{16^k}\tag{1},$$ and with Stirling's formula as $k$ tends to infinity conclude $S={16\over \pi}-4$.

Answer 2

Take the generating function $g(x)=\sum_{n\ge 0}x^n C_n^2$ from https://oeis.org/A001246, note that the sum is given by setting $x=1 /16$ in the generating function, and use the standard properties of the Complete Elliptic Integrals.