Jonathan Miles

2022-07-01

Proving a binomial sum identity $\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{\left(-1{\right)}^{k}}{2k+1}=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}.$

Sanaa Hinton

Expert

You could consider the integral
${\int }_{0}^{1}\left(1-{x}^{2}{\right)}^{n}dx.$

EnvivyEvoxys6

Expert

${S}_{n}=\sum _{k=0}^{n}\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right)}{2k+1}$
We have:
$\begin{array}{rl}{S}_{n}& =\frac{\left(-1{\right)}^{n}}{2n+1}+\sum _{k=0}^{n-1}\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right)}{2k+1}\\ & =\frac{\left(-1{\right)}^{n}}{2n+1}+\sum _{k=0}^{n-1}\left[\frac{n}{n-k}.\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}\right]\\ & =\frac{\left(-1{\right)}^{n}}{2n+1}+\frac{1}{2n+1}\sum _{k=0}^{n-1}\left[\frac{n\left(2n+1\right)}{n-k}.\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}\right]\\ & =\frac{\left(-1{\right)}^{n}}{2n+1}+\frac{1}{2n+1}\sum _{k=0}^{n-1}\left[\frac{2{n}^{2}-2nk+2nk+n}{n-k}.\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}\right]\\ & =\frac{\left(-1{\right)}^{n}}{2n+1}+\frac{1}{2n+1}\sum _{k=0}^{n-1}\left[\frac{2{n}^{2}-2nk}{n-k}.\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}\right]\\ & +\frac{1}{2n+1}\sum _{k=0}^{n-1}\left[\frac{2nk+n}{n-k}.\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}\right]\\ & =\frac{\left(-1{\right)}^{n}}{2n+1}+\frac{2n}{2n+1}\sum _{k=0}^{n-1}\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}+\frac{1}{2n+1}\sum _{k=0}^{n-1}\frac{n\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{n-k}\\ & =\frac{2n}{2n+1}\sum _{k=0}^{n-1}\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}+\frac{1}{2n+1}\sum _{k=0}^{n-1}\left[\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right)\right]+\frac{\left(-1{\right)}^{n}}{2n+1}\\ & =\frac{2n}{2n+1}\sum _{k=0}^{n-1}\frac{\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n-1}{k}\right)}{\left(2k+1\right)}+\frac{1}{2n+1}\sum _{k=0}^{n}\left[\left(-1{\right)}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right)\right]\end{array}$
Therefore,
${S}_{n}=\frac{2n}{2n+1}{S}_{n-1}+0⇒{S}_{n-1}=\frac{2n-2}{2n-1}{S}_{n-2}...⇒{S}_{1}=\frac{2}{3}{S}_{0}$ and ${S}_{0}=1$
Hence,
${S}_{n}=\frac{2n\left(2n-2\right)...2}{\left(2n+1\right)\left(2n-1\right)...3.1}=\frac{\left(2n\right)!!}{\left(2n+1\right)!!}$

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