Power series with cubic factorials

Question

I am looking for a starting point to obtain the closed form of $$ \sum_{k=0}^{\infty} x^k \frac{(2k)!}{(k!)^3}. $$ I have tried transforming the ratio of factorials into combinations, but this lead me nowhere. I also computed several values of the series on Wolfram and I expect the answer to look like $e^{2x} I_0(2x)$ for $I_0$ the $0$-th modified Bessel function. Does anyone has an idea on how to properly obtain the result ?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$

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\begin{align} &\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}x^{k}\, {\pars{2k}! \over \pars{k!}^{3}}} = \sum_{k = 0}^{\infty}{x^{k} \over k!}{2k \choose k} \\[5mm] = &\ \sum_{k = 0}^{\infty}{x^{k} \over k!}\bracks{z^{k}} \pars{1 + z}^{2k} \\[5mm] = &\ \bracks{z^{0}}\sum_{k = 0}^{\infty} {\bracks{{\pars{1 + z}^{2}x/z}}^{k} \over k!} \\[5mm] = &\ \bracks{z^{0}} \exp\pars{{\pars{1 + z}^{2}\,x \over z}} \\[5mm] = &\ \bracks{z^{0}} \exp\pars{x\,\bracks{{1 \over z} + 2 + z}} \\[5mm] = &\ \expo{2x}\bracks{z^{0}} \exp\pars{{1 \over 2}\bracks{\color{red}{2x}}\,\bracks{z + z^{-1}}} \end{align}

The last exponential in the right hand side is the Beseel $\ds{\on{I}_{\nu}}$ Generating Function. Namely, \begin{align} &\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}x^{k}\, {\pars{2k}! \over \pars{k!}^{3}}} = \bbx{\expo{2x}\on{I}_{0}\pars{2x}} \\ & \end{align}

Answer 2

Yes, that is the correct formula for the exponential generating function of the central binomial coefficients $\binom{2k}{k}$. See https://oeis.org/A000984

Answer 3

CONJECTURE: $$\sum_{k=0}^\infty x^k\frac{(2k)!}{k!^3}=e^{2x}I_0(2x)$$ Here's my attempt at showing this using a Cauchy product -

The power series for $e^{2x}$ is: $$e^{2x}=\sum_{k=0}^\infty \frac{2^k}{k!}x^k=\sum_{k=0}^\infty a_k x^k$$ And a well known expansion for $I_0$ is $$I_0(z)=\sum_{k=0}^\infty \frac{z^{2k}}{4^k k!^2}$$ So then $$I_0(2x)=\sum_{k=0}^\infty\frac{x^{2k}}{k!^2}=\sum_{k=0}^\infty b_kx^k$$

Here we have $$b_k=\frac{1+(-1)^{k}}{2}\frac{1}{(k/2)!^2}$$

Then, the Cauchy Product gives $$e^{2x}I_0(2x)=\left(\sum_{k=0}^\infty a_k x^k\right)\cdot\left(\sum_{k=0}^\infty b_kx^k\right)=\sum_{k=0}^\infty c_kx^k$$

$$c_k=\sum_{j=0}^k a_jb_{k-j}=\sum_{j=0}^k b_ja_{k-j}$$ $$=\sum_{j=0}^k\frac{1+(-1)^{n-j}}{2}\frac{1}{((n-j)/2)!^2}\frac{2^j}{j!}=\sum_{j=0}^k \frac{1+(-1)^{j}}{2}\frac{1}{(j/2)!^2}\frac{2^{k-j}}{(k-j)!}$$ I suppose all there is left to do now is show the above are equal to $(2k)!/k!^3$. (sum experts, help!) This is confirmed by Mathematica.

And of course the half integer Gamma is well known. We then have $$=\frac{4^k(2k)!\sqrt{\pi}}{\sqrt{\pi}4^k k!(k!)^2}=\frac{(2k)!}{k!^3}$$ However I think I'm a bit out of my depth to show this using "pencil and paper" methods.