Using Taylor expansion to evaluate infinite sum

Question

How do I use the Taylor expansion of $$(1+x)^{-\frac{1}{2}} $$ to evaluate

$$ \sum_{n=0}^{\infty}\binom{2n}{n}\left(-\dfrac{6}{25}\right)^{n} $$

Thanks

Answer 1

HINT:

$$\binom{2n}n\left(-\frac6{25}\right)^n=\frac{1\cdot2\cdots(2n-1)2n}{1\cdot2\cdots n}\left(-\frac6{25}\right)^n$$

$$=\frac{1\cdot3\cdots(2n-1)\cdot2\cdot4\cdots(2n)}{1\cdot2\cdots n}\left(-\frac6{25}\right)^n$$

$$=1\cdot3\cdots(2n-1)\cdot2^n\left(-\frac6{25}\right)^n$$

$$=2^n(-2)^n\frac{\left(-\frac12\right)\left(-\frac12-1\right)\cdots \left(-\frac12-(n-1)\right)}{n!}\left(-\frac6{25}\right)^n$$

$$= \frac{\left(-\frac12\right)\left(-\frac12-1\right)\cdots \left(-\frac12-(n-1)\right)}{n!}\left(\frac{24}{25}\right)^n$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\mbox{Let's consider}\ \fermi\pars{x} \equiv \sum_{n = 0}^{\infty}{2n \choose n}x^{n}\quad\mbox{where}\quad \fermi\pars{0} = 1.\quad \mbox{Then}, \sum_{n = 0}^{\infty}{2n \choose n}\pars{-\,{6 \over 25}}^{n} = \fermi\pars{-\,{6 \over 25}}}$

\begin{align} \fermi'\pars{x}&=\sum_{n = 1}^{\infty}{\pars{2n}! \over \pars{n - 1}!\,n!}x^{n - 1} =\sum_{n = 0}^{\infty}{\pars{2n + 2}! \over n!\,\pars{n + 1}!}x^{n} =2\sum_{n = 0}^{\infty}\pars{2n + 1}{\pars{2n}! \over n!\,n!}x^{n} \\[3mm]&=4x\fermi'\pars{x} + 2\,\fermi\pars{x} \quad\imp\quad\color{#c00000}{\fermi'\pars{x} + {2 \over 4x - 1}\,\fermi\pars{x} = 0\,,\quad\fermi\pars{0} = 1} \end{align}

$$ \mbox{The solution of the}\ \color{#c00000}{red}\ \mbox{equation is trivial:}\ \fermi\pars{x} = {1 \over \root{1 - 4x}} $$

Then $$ \color{#00f}{\large\sum_{n = 0}^{\infty}{2n \choose n}\pars{-\,{6 \over 25}}^{n}} =\fermi\pars{-\,{6 \over 25}}={1 \over \root{1 - 4\pars{-6/25}}} = \color{#00f}{\large{5 \over 7}} $$

Answer 3

The hardest part of this question is noticing that $\displaystyle(2n-1)!!=\frac{(2n)!}{2^nn!}$. This can be proven quite easily.

Now just work out the general term and substitute in an appropriate value for x.

Note that $(2n-1)!!=(2n-1)\times(2n-3)\times\cdots\times5\times3\times1$.