Two factorials series convergence

Question

How to check whether $$\sum_n \frac{(-1)^n (2n)!x^n}{(n!)^2 4^n}$$ converges or diverges. The ratio test is inconclusive and I don't know what to do with $(2n)!$

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 0}^{\infty}{\pars{-1}^{n}\pars{2n}!\, x^{n} \over \pars{n!}^{2}\, 4^{n}} & = \sum_{n = 0}^{\infty}{2n \choose n}\pars{-\,{x \over 4}}^{n} = \sum_{n = 0}^{\infty}\bracks{{-1/2 \choose n}\pars{-4}^{n}} \pars{-\,{x \over 4}}^{n} \\[5mm] & = \sum_{n = 0}^{\infty}{-1/2 \choose n}x^{n} = \pars{1 + x}^{-1/2} = \bbx{1 \over \root{1 + x}} \\[5mm] & \mbox{} \end{align}

See this post.

Note that $\ds{{-1/2 \choose n} \sim {n^{-1/2} \over \root{\pi}}}$ as $\ds{n \to \infty}$.