Mylee Underwood

Answered

2022-07-01

Denote the sum
${S}_{n}:=\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{\left(-1{\right)}^{n-k}}{n+k+1}$

Answer & Explanation

Camron Herrera

Expert

2022-07-02Added 16 answers

We have

where in step $\left(\star \right)$ we use the substitution $k\to n-k$

Kristen Stokes

Expert

2022-07-03Added 2 answers

To evaluate the sum
$\sum _{k=0}^{n}\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{\left(-1{\right)}^{n-k}}{n+k+1}$
introduce the function
$f\left(z\right)=\frac{n!}{n+1+z}\prod _{q=0}^{n}\frac{1}{z-q}$
which has the property that for $0\le k\le n$
${\mathrm{R}\mathrm{e}\mathrm{s}}_{z=k}f\left(z\right)=\frac{n!}{n+1+k}\prod _{q=0}^{k-1}\frac{1}{k-q}\prod _{q=k+1}^{n}\frac{1}{k-q}\phantom{\rule{0ex}{0ex}}=\frac{n!}{n+1+k}\frac{1}{k!}\frac{\left(-1{\right)}^{n-k}}{\left(n-k\right)!}=\left(\genfrac{}{}{0}{}{n}{k}\right)\frac{\left(-1{\right)}^{n-k}}{n+1+k}.$
Now with residues summing to zero and the residue at infinity being zero by inspection we get for our sum the value
$-{\mathrm{R}\mathrm{e}\mathrm{s}}_{z=-n-1}f\left(z\right)=-n!\prod _{q=0}^{n}\frac{1}{-n-1-q}\phantom{\rule{0ex}{0ex}}=n!\left(-1{\right)}^{n}\prod _{q=0}^{n}\frac{1}{n+1+q}=\left(-1{\right)}^{n}n!\frac{n!}{\left(2n+1\right)!}.$
This is
$\frac{\left(-1{\right)}^{n}}{2n+1}{\left(\genfrac{}{}{0}{}{2n}{n}\right)}^{-1}.$

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