How to prove $\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2} = 2\sqrt{2} \;$?

Question

I found out that the sum $$\sum_{n=0}^{\infty} \frac {(2n+1)!} {2^{3n} \; (n!)^2}$$ converges to $2\sqrt{2}$.

But right now I don't have enough time to figure out how to solve this. I would really appreciate any help. Just one tiny hint might help too.

Answer 1

Hint: Observe that (by the binomial theorem for instance) $$ \sum_{n=0}^{\infty} \frac {(2n)!} {(n!)^2}x^{2n}=\frac{1}{\sqrt{1-4x^2}}, \quad |x|<\frac{1}{2}. $$ Then multiply by $x$ and perform a termwise differentiation and you readily obtain the desired result with $x^2=\dfrac{1}{2^3}.$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{\verts{\mu} < {1 \over 4}}$:

\begin{align} \color{#66f}{\large\sum_{n\ =\ 0}^{\infty}\mu^{n}{\pars{2n + 1}! \over \pars{n!}^{2}}} &=\sum_{n\ =\ 0}^{\infty}\pars{2n + 1}\mu^{n}{2n \choose n} =\pars{2\mu\,\totald{}{\mu} + 1} \color{#c00000}{\sum_{n\ =\ 0}^{\infty}\mu^{n}{2n \choose n}} \end{align}

\begin{align} \color{#c00000}{\sum_{n\ =\ 0}^{\infty}\mu^{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} =\oint_{\verts{z}\ =\ 1}{1 \over z} \sum_{n\ =\ 0}^{\infty}\bracks{\pars{1 + z}^{2}\mu \over z}^{n} \,{\dd z \over 2\pi\ic} \\[5mm]&=\oint_{\verts{z}\ =\ 1} {1 \over z}{1 \over 1 - \pars{1 + z}^{2}\mu/z}\,{\dd z \over 2\pi\ic} =-\oint_{\verts{z}\ =\ 1}{1 \over \mu z^{2} + \pars{2\mu - 1}z + \mu} \,{\dd z \over 2\pi\ic} \\[3mm]&=-\oint_{\verts{z}\ =\ 1}{1 \over \mu\pars{z - r_{-}}\pars{z - r_{+}}} \,{\dd z \over 2\pi\ic} \\[5mm]&\mbox{where}\quad \boxed{\ds{\quad r_{\pm} \equiv {1 - 2\mu \pm \root{1 - 4\mu} \over 2\mu}\quad}} \end{align}

Note that $\ds{\verts{r_{-}}\ <\ 1}$ and $\ds{\verts{r_{+}}\ >\ 1}$ such that:

\begin{align} \color{#c00000}{\sum_{n\ =\ 0}^{\infty}\mu^{n}{2n \choose n}} &=-\,{1 \over \mu}\,{1 \over r_{-} - r_{+}} =-\,{1 \over \mu}{1 \over -2\root{1 - 4\mu}/\pars{2\mu}} ={1 \over \root{1 - 4\mu}} \end{align}

Then, \begin{align} \color{#66f}{\large\sum_{n\ =\ 0}^{\infty}\mu^{n}{\pars{2n + 1}! \over \pars{n!}^{2}}} &=\pars{2\mu\,\totald{}{\mu} + 1}{1 \over \root{1 - 4\mu}} =\color{#66f}{\large{1 \over \root{1 - 4\mu}} + {4\mu \over \pars{1 - 4\mu}^{3/2}}} \end{align}

Set $\ds{\mu = {1 \over 8}}$ in both members: \begin{align} \color{#66f}{\large% \sum_{n\ =\ 0}^{\infty}{\pars{2n + 1}! \over 2^{3n}\pars{n!}^{2}}} &=\color{#66f}{\large2\root{2}} \approx {\tt 2.8284} \end{align}