Mylee Underwood

Answered

2022-07-01

Denote the sum

${S}_{n}:=\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{n+k+1}$

${S}_{n}:=\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{n+k+1}$

Answer & Explanation

Camron Herrera

Expert

2022-07-02Added 16 answers

We have

$\begin{array}{rl}{\int}_{0}^{1}(x(1-x){)}^{n}\text{}\text{dx}& ={\int}_{0}^{1}(x-{x}^{2}{)}^{n}\text{}\text{dx}\\ & ={\int}_{0}^{1}\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{x}^{k}(-{x}^{2}{)}^{n-k}\text{}\text{dx}\\ & =\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}(-1{)}^{n-k}{\int}_{0}^{1}{x}^{2n-k}\text{}\text{dx}\\ & =\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{2n-k+1}\\ & \stackrel{\star}{=}\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{n-k}{\textstyle )}\frac{(-1{)}^{k}}{2n-(n-k)+1}\\ & =\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{k}}{n+k+1}.\end{array}$

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

$\begin{array}{rl}{\int}_{0}^{1}(x(1-x){)}^{n}\text{}\text{dx}& ={\int}_{0}^{1}(x-{x}^{2}{)}^{n}\text{}\text{dx}\\ & ={\int}_{0}^{1}\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{x}^{k}(-{x}^{2}{)}^{n-k}\text{}\text{dx}\\ & =\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}(-1{)}^{n-k}{\int}_{0}^{1}{x}^{2n-k}\text{}\text{dx}\\ & =\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{2n-k+1}\\ & \stackrel{\star}{=}\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{n-k}{\textstyle )}\frac{(-1{)}^{k}}{2n-(n-k)+1}\\ & =\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{k}}{n+k+1}.\end{array}$

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

Kristen Stokes

Expert

2022-07-03Added 2 answers

To evaluate the sum

$\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{n+k+1}$

introduce the function

$f(z)=\frac{n!}{n+1+z}\prod _{q=0}^{n}\frac{1}{z-q}$

which has the property that for $0\le k\le n$

${\mathrm{R}\mathrm{e}\mathrm{s}}_{z=k}f(z)=\frac{n!}{n+1+k}\prod _{q=0}^{k-1}\frac{1}{k-q}\prod _{q=k+1}^{n}\frac{1}{k-q}\phantom{\rule{0ex}{0ex}}=\frac{n!}{n+1+k}\frac{1}{k!}\frac{(-1{)}^{n-k}}{(n-k)!}={\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{n+1+k}.$

Now with residues summing to zero and the residue at infinity being zero by inspection we get for our sum the value

$-{\mathrm{R}\mathrm{e}\mathrm{s}}_{z=-n-1}f(z)=-n!\prod _{q=0}^{n}\frac{1}{-n-1-q}\phantom{\rule{0ex}{0ex}}=n!(-1{)}^{n}\prod _{q=0}^{n}\frac{1}{n+1+q}=(-1{)}^{n}n!\frac{n!}{(2n+1)!}.$

This is

$\frac{(-1{)}^{n}}{2n+1}{{\textstyle (}\genfrac{}{}{0ex}{}{2n}{n}{\textstyle )}}^{-1}.$

$\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{n+k+1}$

introduce the function

$f(z)=\frac{n!}{n+1+z}\prod _{q=0}^{n}\frac{1}{z-q}$

which has the property that for $0\le k\le n$

${\mathrm{R}\mathrm{e}\mathrm{s}}_{z=k}f(z)=\frac{n!}{n+1+k}\prod _{q=0}^{k-1}\frac{1}{k-q}\prod _{q=k+1}^{n}\frac{1}{k-q}\phantom{\rule{0ex}{0ex}}=\frac{n!}{n+1+k}\frac{1}{k!}\frac{(-1{)}^{n-k}}{(n-k)!}={\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}\frac{(-1{)}^{n-k}}{n+1+k}.$

Now with residues summing to zero and the residue at infinity being zero by inspection we get for our sum the value

$-{\mathrm{R}\mathrm{e}\mathrm{s}}_{z=-n-1}f(z)=-n!\prod _{q=0}^{n}\frac{1}{-n-1-q}\phantom{\rule{0ex}{0ex}}=n!(-1{)}^{n}\prod _{q=0}^{n}\frac{1}{n+1+q}=(-1{)}^{n}n!\frac{n!}{(2n+1)!}.$

This is

$\frac{(-1{)}^{n}}{2n+1}{{\textstyle (}\genfrac{}{}{0ex}{}{2n}{n}{\textstyle )}}^{-1}.$

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