What is the value of $\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}$ What is the value of $$\sum_{n=0}^{\infty}\left(-\frac{1}{8}\right)^n\binom{2n}{n}\;?$$
EDIT
I bumped into this series when inserting $\overrightarrow{r_1}=\left(\begin{array} {c}0\\0\\1\end{array}\right)$ and $\overrightarrow{r}=\left(\begin{array} {c}1\\1\\0\end{array}\right)$ into $$\frac{1}{|\overrightarrow{r}-\overrightarrow{r_1}|}=\sum_{l=0}^\infty\frac{r_{<}^l}{r_{>}^{l+1}}P_{l}(cos\theta)\;.$$
See (Series representation of $1/|x-x'|$ using legendre polynomials).
So I knew the result, $\sqrt{\frac{2}{3}}$, but wished to find out what other approaches there would be to evaluate the series.
It transpires that I overlooked the relatively standard evaluation of $\sum_{n=0}^{\infty}\binom{2n}{n}x^n$ as being equal to $\sqrt{\frac{1}{1-4x}}$, following a calculation similar to that given by achille-hui below, which requires some complex function theory, in particular when proving that $\binom{2n}{n} = \frac{1}{2\pi}\int_0^{2\pi}\left(e^{i\theta}+e^{-i\theta}\right)^{2n}d\theta$, or that of alex.jordan below, requiring no more than Taylor expansion.

$$
\begin{align}
\sum\binom{2n}{n}x^n
&=\sum\frac{1}{n!}\frac{(2n)!}{n!}x^n\\
&=\sum\frac{1}{n!}2^n(2n-1)(2n-3)\cdots(3)(1)x^n\\
&=\sum\frac{1}{n!}\left(\frac{2n-1}2\right)\left(\frac{2n-3}2\right)\cdots\left(\frac32\right)\left(\frac12\right)(4x)^n\\
&=\sum\frac{1}{n!}\left(-{\frac{2n-1}2}\right)\left(-{\frac{2n-3}2}\right)\cdots\left(-{\frac32}\right)\left(-{\frac12}\right)(-4x)^n\\
&=\sum\frac{1}{n!}f^{(n)}(0)(-4x)^n\\
\end{align}
$$
where $f(z)=(z+1)^{-1/2}$. Now interpret as a Taylor series and evaluate at $x=-{\frac18}$ (using the corresponding $z=\frac12$).
