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I came across the following summation: $$ \sum_{k=0}^n\frac{(-1)^k(2k)!!}{(2k+1)!!}\dbinom{n}{k}\,\,\,\,(n\in\mathbb{N}). $$ $\tbinom{n}{k}$ are binomial coefficients, $n!/k!(n-k)!$.

Mathematica told me that the answer is $\frac{1}{2n+1}$ at least for $n\le20$, but I couldn't prove it.

Does anyone have an idea to prove, for any $n$, $$ \sum_{k=0}^n\frac{(-1)^k(2k)!!}{(2k+1)!!}\dbinom{n}{k}=\frac{1}{2n+1}\,\,\,\,? $$ Thanks in advance.

Yutaro
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2 Answers2

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A way to prove it is using the binomial inversion:

Theorem: If $$a_{n}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}b_{k} $$ then $$b_{n}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}a_{k}.$$

Now since holds $$\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{1}{2k+1}=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}$$ (see for example here) your claim follows. Another way is observing that $$\frac{\left(2k\right)!!}{\left(2k+1\right)!!}=\int_{0}^{1}\left(1-t^{2}\right)^{k}dt $$ hence $$\begin{align} \sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{\left(2k\right)!!}{\left(2k+1\right)!!}= & \int_{0}^{1}\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(1-t^{2}\right)^{k}dt \\ = & \int_{0}^{1}t^{2n}dt \\ = & \color{red}{\frac{1}{2n+1}}. \end{align}$$

Community
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Marco Cantarini
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Since $\ds{\pars{2k}!! = 2^{k}\pars{k!} = 2^{k}\,\Gamma\pars{k + 1}}$ and $\ds{\pars{2k +1}!! = {\pars{2k + 2}! \over 2^{k + 1}\pars{k + 1}!} = 2^{-k - 1}\,\,{\Gamma\pars{2k + 3} \over \Gamma\pars{k + 2}}}$, we'll have: $$ {\pars{2k}!! \over \pars{2k + 1}!!} = 2^{2k + 1}\,\,\,{\Gamma\pars{k + 1}\Gamma\pars{k + 2} \over \Gamma\pars{2k + 3}} = 2^{2k + 1}\int_{0}^{1}t^{k}\pars{1 - t}^{k + 1}\,\dd t $$

and \begin{align} &\color{#f00}{\sum_{k = 0}^{n}\pars{-1}^{k}\, {\pars{2k}!! \over \pars{2k + 1}!!}\,{n \choose k}} = 2\int_{0}^{1}\pars{1 - t} \sum_{k = 0}^{n}{n \choose k}\bracks{-4t\pars{1 - t}}^{\, k}\,\,\dd t \\[5mm] = &\ 2\int_{0}^{1}\pars{1 - t}\bracks{1 - 4t\pars{1 - t}}^{\, n}\,\,\dd t = \int_{0}^{1}\bracks{1 - 4t\pars{1 - t}}^{\, n}\,\,\dd t = \color{#f00}{1 \over 2n + 1} \end{align}

Note that $\ds{1 - 4t\pars{1 - t} = 4\pars{t - \half}^{2}}$.

Felix Marin
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