What is the value of $\sum_{n=0}^{\infty}(-\frac{1}{8})^n\binom{2n}{n}$

Question

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.

Answer 1

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

$$\frac{2}{\pi}\int_0^\infty \frac{1}{1 + \frac{1}{2(1+t^2)}}\frac{dt}{1+t^2} = \frac{2}{\pi}\int_0^\infty \frac{dt}{t^2 + \frac32} = \frac{2}{\pi}\sqrt{\frac23}\left[ \tan^{-1}\sqrt{\frac23} t \right]_0^\infty = \sqrt{\frac23}$$

Answer 2

$$ \begin{align} \sum\binom{2n}{n}x^n &=\sum\frac{1}{n!}\frac{(2n)!}{n!}x^n\\ &=\sum\frac{1}{n!}2^n(2n-1)(2n-3)\cdots(3)(1)x^n\\ &=\sum\frac{1}{n!}\left(\frac{2n-1}2\right)\left(\frac{2n-3}2\right)\cdots\left(\frac32\right)\left(\frac12\right)(4x)^n\\ &=\sum\frac{1}{n!}\left(-{\frac{2n-1}2}\right)\left(-{\frac{2n-3}2}\right)\cdots\left(-{\frac32}\right)\left(-{\frac12}\right)(-4x)^n\\ &=\sum\frac{1}{n!}f^{(n)}(0)(-4x)^n\\ \end{align} $$

where $f(z)=(z+1)^{-1/2}$. Now interpret as a Taylor series and evaluate at $x=-{\frac18}$ (using the corresponding $z=\frac12$).

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\sum_{n\ =\ 0}^{\infty}\pars{-\,{1 \over 8}}^{n}{2n \choose n}:\ {\large ?}}$.

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}