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I'm trying to solve one of my combinatorics exercise but I struggle a bit.

Is the equality correct for all the $n\ge 0$? $$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$$

First of all:$$\sum_{n=0}^\infty \left(2^{-2n}\binom{2n}{n}\right)x^n=\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$$ Now

$$\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k \right)x^n=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=k}^\infty \binom{2k}{k}\binom{n}{k} x^n=[n-k=m]=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{m=0}^\infty \binom{2k}{k}\binom{m+k}{m} x^{m+k}=\sum_{k=0}^\infty \left(-\frac{x}{4}\right)^k\binom{2k}{k}\sum_{m=0}^\infty \binom{m+k}{m} x^{m}$$ and here I don't know what to do next. Can anyone help me? Thanks in advice!

Karagum
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3 Answers3

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You may consider that $$ \frac{1}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2k}(\theta)\,d\theta \tag{A}$$ from which it follows that: $$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k=0}^{n}\binom{n}{k}(-\sin^2\theta)^k =\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta\tag{B}$$ and the conclusion is straightforward through the substitution $\theta\mapsto\frac{\pi}{2}-\varphi$.

With generating functions, from $$ \sum_{k\geq 0}\binom{2k}{k}\frac{z^k}{4^k}=\frac{1}{\sqrt{1-z}} \tag{C}$$ by replacing $z$ with $\frac{x}{1+x}$, then by multiplying both sides by $\frac{1}{1+x}$, we get $$ \sum_{k\geq 0}\binom{2k}{k}\frac{x^k}{4^k(1+x)^{k+1}} = \frac{1}{\sqrt{1-x}}\tag{D}$$ then by applying $[x^n]$ to both sides: $$ \sum_{k\geq 0}\binom{2k}{k}\frac{(-1)^k}{4^k}\binom{n}{k} = \frac{1}{4^n}\binom{2n}{n}\tag{E}$$ where the RHS has been managed through $(C)$ and the LHS through stars and bars.

Jack D'Aurizio
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This follows directly (more or less) from Vandermonde's identity: \begin{align*} \sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} &= \binom{n-1/2}{n} \\ &= \frac{(n-1/2)(n-3/2)\cdots(n-1/2-(n-1))}{n!}\\ &= \frac{(2n-1)(2n-3)\cdots (1)}{2^nn!} \\ &= 2^{-2n}\binom{2n}{n}. \end{align*} But also, $$\binom{-1/2}{k} = \frac{(-1/2)(-1/2-1)\cdots(-1/2-(k-1)}{k!} = (-1)^k\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2^kk!} = (-1)^k 2^{-2k}\binom{2k}{k}$$ and therefore $$\sum_{k=0}^n \binom{n}{k}\binom{-1/2}{k} = \sum_{k=0}^n \binom{n}{k}(-1)^k2^{-2k}\binom{2k}{k}.$$

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

In 0ne of my previous answers I already showed that \begin{equation} {2m \choose m} = {-1/2 \choose m}\pars{-4}^{m}\label{1}\tag{1} \end{equation}

\begin{align} \mbox{So,}\quad\sum_{k = 0}^{\infty}{2k \choose k}{n \choose k}\pars{-\,{1 \over 4}}^{k} & = \sum_{k = 0}^{\infty}{n \choose k}\pars{-\,{1 \over 4}}^{k}\ \overbrace{\bracks{{-1/2 \choose k}\pars{-4}^{k}}}^{\ds{2k \choose k}} \\[5mm] & = \sum_{k = 0}^{\infty}{n \choose k}\ \overbrace{\bracks{z^{k}}\pars{1 + z}^{-1/2}}^{\ds{-1/2 \choose k}}\ =\ \bracks{z^{0}}\pars{1 + z}^{-1/2}\sum_{k = 0}^{\infty}{n \choose k} \pars{1 \over z}^{k} \\[5mm] & = \bracks{z^{0}}\pars{1 + z}^{-1/2}\bracks{\pars{1 + {1 \over z}}^{n}} = \bracks{z^{n}}\pars{1 + z}^{n - 1/2} \\[5mm] & = {n - 1/2 \choose n} = {-1/2 \choose n}\pars{-1}^{n} = \pars{1 \over 4}^{n}\ \overbrace{\bracks{{-1/2 \choose n}\pars{-4}^{n}}}^{\ds{{2n \choose n}.\ \mbox{See}\ \eqref{1}}} \\[5mm] & = \bbx{2^{-2n}{2n \choose n}} \end{align}

Felix Marin
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