Convergence and value of a binomial series

Question

I wanted to know whether I am allowed to write:

\begin{equation} \sum_{n=0}^{\infty} \binom{2n}{n} a^{-n} = (1+a^{-1})^{2n}, \ \mathrm{for} \ n > 4? \end{equation}

I was wondering whether the exponent ($2n$ in this case) must be independent of the index of the series (i.e. $n$). Thanks.

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Answer 1

The correct identity is $$ \forall|a|<\frac{1}{4},\qquad\sum_{n\geq 0}\binom{2n}{n}a^n = \frac{1}{\sqrt{1-4a}} $$ and it can be proved through the extended binomial theorem or the fact that $$ \frac{1}{4^n}\binom{2n}{n} = \frac{1}{\pi}\int_{0}^{\pi}\sin(x)^{2n}\,dx.$$

Answer 2

No, your equation cannot be correct, because the LHS is a summation over the index variable $n$; thus the RHS cannot be a function of $n$.

Answer 3

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{f}\pars{x} \equiv \sum_{n = 0}^{\infty}{2n \choose n}x^{n}:\ ?\,,\qquad\mrm{f}\pars{0} = 1\,, \qquad \verts{x} < {1 \over 4}}$.

\begin{align} \mrm{f}'\pars{x} & = \sum_{n = 1}^{\infty}{\pars{2n}! \over n!\pars{n - 1}!}x^{n - 1} = \sum_{n = 0}^{\infty}{\pars{2n + 2}! \over \pars{n + 1}!\, n!}x^{n} = 2\sum_{n = 0}^{\infty}\pars{2n + 1}{\pars{2n}! \over n!\, n!}x^{n} \\[5mm] & = 4x\,\totald{}{x}\sum_{n = 0}^{\infty}{2n \choose n}\,x^{n} + 2\sum_{n = 0}^{\infty}{2n \choose n}\,x^{n} = 4x\,\mrm{f}'\pars{x} + 2\mrm{f}\pars{x} \\[5mm] & \implies \bbox[8px,border:1px groove navy]{\mrm{f}'\pars{x} + {2 \over 4x - 1}\,\mrm{f}\pars{x} = 0} \end{align} This differential equation can be rewritten as $\ds{\pars{~\mbox{with}\ \verts{x} < 1/4~}}$ $$ \totald{\bracks{\root{1 - 4x}\,\mrm{f}\pars{x}}}{x} = 0 \implies \root{1 - 4x}\,\mrm{f}\pars{x} = \root{1 - 4 \times 0}\ \overbrace{\,\mrm{f}\pars{0}}^{\ds{=\ 1}} = 1 $$

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$$ \mrm{f}\pars{x} \equiv \sum_{n = 0}^{\infty}{2n \choose n}x^{n} = \bbox[8px,border:1px groove navy]{{1 \over \root{1 - 4x}}\,, \qquad\verts{x} < {1 \over 4}} $$

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$$ \sum_{n = 0}^{\infty}{2n \choose n}a^{-n} = \bbox[8px,border:1px groove navy]{\root{a \over a - 4}\,, \qquad\verts{a} > 4} $$