So the question is -
$\displaystyle S = \sum_{n=1}^\infty{\frac{1}{10^n}\left(\begin{matrix}2n\\ n\end{matrix}\right)}$. Find $S$.
I tried converting the $n^{th}$ term as a difference of two terms but that didn't worked quite well. I am clueless and don't know how to proceed.
From the OGF of Catalan numbers we have that: $$ \sum_{n\geq 0}\binom{2n}{n}z^n = \frac{1}{\sqrt{1-4z}} \tag{1}$$ for any $z\in\mathbb{C}$ such that $|z|<\frac{1}{4}$. By substituting $z=\frac{1}{10}$: $$ \sum_{n\geq 1}\binom{2n}{n}\frac{1}{10^n}=-1+\frac{1}{\sqrt{1-4/10}}=\color{red}{\frac{\sqrt{15}-3}{3}}.\tag{2}$$
As an alternative approach, we can prove through integration by parts that: $$ \int_{0}^{\pi}\sin^{2n}\theta\,d\theta = \frac{\pi}{4^n}\binom{2n}{n}, \tag{3}$$ hence: $$\sum_{n\geq 1}\binom{2n}{n}\frac{1}{10^n}=\frac{1}{\pi}\int_{0}^{\pi}\frac{\frac{2}{5}\sin^2\theta}{1-\frac{2}{5}\sin^2\theta}\,d\theta=\frac{4}{5\pi}\int_{0}^{+\infty}\frac{t^2}{(1+t^2)\left(1+\frac{3}{5}t^2\right)}\,dt.\tag{4} $$ Now, by partial fraction decomposition, the last integral equals: $$ \frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{1+\frac{3}{5}t^2}-\frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{1+t^2}=-1+\sqrt{\frac{5}{3}}\tag{5}$$ as wanted. The same chain of identities can be used to prove $(1)$.