Infinite series involving binomial and geometric How to find the following sum:
$$\sum_{x=0}^{\infty}\left(-\frac{1}{16}\right)^x \binom{2x}{x}$$
 A: \begin{eqnarray*}
\binom{2n}{n} = \binom{-1/2}{n}(-4)^{n}.
\end{eqnarray*}
\begin{eqnarray*}
\sum_{n=0}^{\infty} \binom{2n}{n} \left( \frac{-1}{16} \right)^n = \sum_{n=0}^{\infty} \binom{-1/2}{n} \left( \frac{1}{4} \right)^n =\frac{1}{\sqrt{1+\frac{1}{4}}} =\color{red}{\frac{2}{\sqrt{5}}}.
\end{eqnarray*}
A: Due to De Moivre's identity $\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}$ and the orthogonality relation $\int_{0}^{2\pi}e^{ni\theta}e^{-mi\theta}\,d\theta=2\pi \delta(m,n)$ we have the following integral representation for central binomial coefficients:
$$ \frac{1}{4^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta\tag{1}$$
leading to
$$ \sum_{n\geq 0}\frac{(-1)^n}{16^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{n\geq 0}\frac{(-1)^n}{4^n}\left(\cos\theta\right)^{2n}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1+\frac{1}{4}\cos^2\theta}\tag{2} $$
an by enforcing the substitution $\theta=\arctan u$ we get:
$$ \sum_{n\geq 0}\frac{(-1)^n}{16^n}\binom{2n}{n} = \frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{\frac{5}{4}+u^2}=\frac{2}{\sqrt{5}}.\tag{3} $$
The same approach can be used to prove the following identity:
$$ \forall z\in(-1,1),\qquad \sum_{n\geq 0}\frac{z^n}{4^n}\binom{2n}{n}=\frac{1}{\sqrt{1-z}}\tag{4} $$
whose combinatorial equivalent is given by
$$ \sum_{n=0}^{N}\binom{2n}{n}\binom{2N-2n}{N-n}=4^N.\tag{5} $$
