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I want to prove the following summation:

$$\sum_{i = 1}^{k - 1}\frac{(2i)!}{(2^ii!)^2}\cdot\frac{(2(k-i))!}{(2^{k-i}(k-i)!)^2} = 1 - \frac{2(2k)!}{(2^kk!)^2}$$

There are a lot of factorials going on, and I have no idea how to simplify them out or perhaps write them as binomial coefficients somehow. I would appreciate some help or hints about what to try.

b_pcakes
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2 Answers2

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The identity can be written by using binomial coefficients as $$\frac{1}{4^k}\sum_{i=1}^{k-1} \binom{2i}{i}\binom{2(k-i)}{k-i}= 1 - \frac{2}{4^k}\binom{2k}{k}.$$ That is $$\binom{2k}{k}+\sum_{i=1}^{k-1} \binom{2i}{i}\binom{2(k-i)}{k-i} +\binom{2k}{k}=4^k.$$ Now you should recognize the convolution of central binomial coefficients.

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Robert Z
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

\begin{equation} \sum_{i = 1}^{k - 1}{\pars{2i}! \over \pars{2^{i}i!}^{2}}\ {\bracks{2\pars{k - i}}! \over \bracks{2^{k - i}\,\pars{k - i}!}^{2}} = 1 - {2\pars{2k}! \over \pars{2^{k}\,k!}^{2}}:\ {\large ?} \label{1}\tag{1} \end{equation}

\begin{align} &\sum_{i = 1}^{k - 1}{\pars{2i}! \over \pars{2^{i}i!}^{2}}\ {\bracks{2\pars{k - i}}! \over \bracks{2^{k - i}\,\pars{k - i}!}^{2}} = {1 \over 2^{2k}}\sum_{i = 1}^{k - 1}{2i \choose i}{2k - 2i \choose k - i} \\[5mm] = &\ {1 \over 2^{2k}}\bracks{-{2 \times 0 \choose 0}{2k - 2 \times 0 \choose k - 0} + \sum_{i = 0}^{k}{2i \choose i}{2k - 2i \choose k - i} - {2k \choose k}{2k - 2k \choose k - k}} \\[5mm] = &\ {1 \over 2^{2k}}\sum_{i = 0}^{\color{#f00}{\infty}}{2i \choose i}{2k - 2i \choose k - i} - {2 \over 2^{2k}}{2k \choose k} = {1 \over 2^{2k}}\sum_{i = 0}^{\infty}{2i \choose i}{2k - 2i \choose k - i} - {2\pars{2k}! \over \pars{2^{k}\,k!}^{2}} \label{2}\tag{2} \end{align}

From \eqref{1} and \eqref{2}, we see $\underline{\ the\ proof\ will\ be\ complete\ }$ once we show

\begin{equation} \sum_{j = 0}^{\infty}{2j \choose j}{2k - 2j \choose k - j} = \bbx{\ds{2^{2k}}}\label{3}\tag{3} \end{equation}

Then, \begin{align} &\sum_{j = 0}^{\infty}{2j \choose j}{2k - 2j \choose k - j} = \sum_{j = 0}^{\infty}{-1/2 \choose j}\pars{-4}^{\, j} {-1/2 \choose k - j}\pars{-4}^{k - j}\label{4}\tag{4} \end{align} where we used a identity from one of my previous answers. \eqref{4} is reduced to: \begin{align} &\sum_{j = 0}^{\infty}{2j \choose j}{2k - 2j \choose k - j} = \pars{-4}^{k}\sum_{j = 0}^{\infty}{-1/2 \choose j} \oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-1/2} \over z^{k - j + 1}}\, {\dd z \over 2\pi\ic} \\[5mm] = &\ \pars{-4}^{k}\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-1/2} \over z^{k + 1}} \sum_{j = 0}^{\infty}{-1/2 \choose j}z^{\,j}\,{\dd z \over 2\pi\ic} \\[5mm] = &\ \pars{-4}^{k}\oint_{\verts{z} = 1^{-}}{\pars{1 + z}^{-1/2} \over z^{k + 1}}\, \pars{1 + z}^{-1/2}\,\,\,{\dd z \over 2\pi\ic} = \pars{-4}^{k}\ \overbrace{\bracks{z^{k}}\pars{1 \over 1 + z}}^{\ds{\pars{-1}^{k}}}\ =\ \bbx{\ds{2^{2k}}} \end{align}
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Felix Marin
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