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.
 A: As mentioned in Dr. Graubner's comment, the sum is $\sqrt{\frac23}$.
In fact, this is a special case of a sort of famous Taylor series expansion:;
$$\frac{1}{\sqrt{1-4z}} = \sum_{n=0}^\infty \binom{2n}{n} z^n$$
which appeared as limiting example of several theorems in complex analysis.
To compute the series ourselves, we can use following integral representation of 
the binomial coefficients:
$$\binom{2n}{n} = \frac{1}{2\pi}\int_0^{2\pi} (e^{i\theta} + e^{-i\theta})^{2n} d\theta
= \frac{4^n}{2\pi}\int_0^{2\pi} \cos^{2n}\theta d\theta
$$
Substitute this into our sum, we find the sum is equal to
$$
\frac{1}{2\pi}\int_0^{2\pi} \sum_{n=0}^\infty \left(-\frac12\cos^2\theta\right)^n d\theta
= \frac{1}{2\pi}\int_0^{2\pi} \frac{1}{1+\frac12\cos^2\theta} d\theta
= \frac{2}{\pi}\int_0^{\pi/2} \frac{1}{1+\frac12\cos^2\theta} d\theta
$$
Introduce change of variable $t = \tan\theta$, this becomes
$$\frac{2}{\pi}\int_0^\infty \frac{1}{1 + \frac{1}{2(1+t^2)}}\frac{dt}{1+t^2}
= \frac{2}{\pi}\int_0^\infty \frac{dt}{t^2 + \frac32}
= \frac{2}{\pi}\sqrt{\frac23}\left[ \tan^{-1}\sqrt{\frac23} t \right]_0^\infty
= \sqrt{\frac23}$$
A: $$
\begin{align}
\sum\binom{2n}{n}x^n
&=\sum\frac{1}{n!}\frac{(2n)!}{n!}x^n\\
&=\sum\frac{1}{n!}2^n(2n-1)(2n-3)\cdots(3)(1)x^n\\
&=\sum\frac{1}{n!}\left(\frac{2n-1}2\right)\left(\frac{2n-3}2\right)\cdots\left(\frac32\right)\left(\frac12\right)(4x)^n\\
&=\sum\frac{1}{n!}\left(-{\frac{2n-1}2}\right)\left(-{\frac{2n-3}2}\right)\cdots\left(-{\frac32}\right)\left(-{\frac12}\right)(-4x)^n\\
&=\sum\frac{1}{n!}f^{(n)}(0)(-4x)^n\\
\end{align}
$$
where $f(z)=(z+1)^{-1/2}$. Now interpret as a Taylor series and evaluate at $x=-{\frac18}$ (using the corresponding $z=\frac12$).
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\sum_{n\ =\ 0}^{\infty}\pars{-\,{1 \over 8}}^{n}{2n \choose n}:\ {\large ?}}$.

With $\ds{\mu \equiv -\,{1 \over 8}}$:

\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 0}^{\infty}\pars{-\,{1 \over 8}}^{n}{2n \choose n}}
=\sum_{n\ =\ 0}^{\infty}\mu^{n}
\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{2n} \over z^{n + 1}}\,{\dd z \over 2\pi\ic}
\\[5mm]&=\oint_{\verts{z}\ =\ 1}{1 \over z}
\sum_{n\ =\ 0}^{\infty}\bracks{\mu\pars{1 + z}^{2} \over z}^{n}
\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}{1 \over z}{1 \over 1 + \pars{1 + z}^{2}/\pars{8z}}
\,{\dd z \over 2\pi\ic}
\\[5mm]&=8\oint_{\verts{z}\ =\ 1}{1 \over z^{2} + 10z + 1}\,{\dd z \over 2\pi\ic}
=\left. 8\,{1 \over 2z + 10}\right\vert_{\,z\ =\ 2\root{6}\ -\ 5}
={4 \over 2\root{6}}=\color{#66f}{\large\root{2 \over 3}}
\end{align}
