siutulysr5

Answered

2022-01-23

Evaluate

$\sum _{n,k}\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}$

where the summation ranges over all positive integers n, k with$1<k<n-1$

where the summation ranges over all positive integers n, k with

Answer & Explanation

Frauental91

Expert

2022-01-24Added 15 answers

Let $l=n-k$ , we have

$\sum _{n=4}^{\mathrm{\infty}}\sum _{k=2}^{n-2}{\left(\begin{array}{c}n\\ k\end{array}\right)}^{-1}=\sum _{k=2}^{\mathrm{\infty}}\sum _{n=k+2}^{\mathrm{\infty}}{\left(\begin{array}{c}n\\ k\end{array}\right)}^{-1}$

$=\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}{\left(\begin{array}{c}k+l\\ k\end{array}\right)}^{-1}=\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}\frac{k!l!}{(k+l)!}$

$=\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}(k+l+1)\frac{\mathrm{\Gamma}(k+1)\mathrm{\Gamma}(l+1)}{\mathrm{\Gamma}(k+l+2)}$

$=\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}(k+l+1){\int}_{0}^{1}{t}^{k}{(1-t)}^{l}dt$

$={\int}_{0}^{1}\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}\left[(k+l+1){t}^{k}{(1-t)}^{l}\right]dt$

Notice when s and t are independent, we have

$\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}(k+l+1){t}^{k}{s}^{l}=\sum _{k=2}^{\mathrm{\infty}}\sum _{l=2}^{\mathrm{\infty}}(t\frac{d}{dt}+s\frac{d}{ds}+1){t}^{k}{s}^{l}$

$=(t\frac{d}{dt}+s\frac{d}{ds}+1)\frac{{t}^{2}{s}^{2}}{(1-t)(1-s)}$

$=\frac{{s}^{2}{t}^{2}(5-4(s+t)+3st)}{{(1-s)}^{}}$

Notice when s and t are independent, we have

0

Aaron Hughes

Expert

2022-01-25Added 13 answers

Partial Fractions and Telescoping Sums

change order of summation

$\sum _{n=4}^{\mathrm{\infty}}\sum _{k=2}^{n-2}\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}=\sum _{k=2}^{\mathrm{\infty}}\sum _{n=k+2}^{\mathrm{\infty}}\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}$

$=\sum _{k=2}^{\mathrm{\infty}}\frac{k}{k-1}\sum _{n=k+2}^{\mathrm{\infty}}(\frac{1}{\left(\begin{array}{c}n-1\\ k-1\end{array}\right)}-\frac{1}{\left(\begin{array}{c}n\\ k-1\end{array}\right)})$

Telescoping sum

$=\sum _{k=2}^{\mathrm{\infty}}\frac{k}{k-1}\frac{1}{\left(\begin{array}{c}k+1\\ k-1\end{array}\right)}$

$=\sum _{k=2}^{\mathrm{\infty}}\frac{2}{(k-1)(k+1)}$

Partial fractions$\sum _{k=2}^{\mathrm{\infty}}(\frac{1}{k-1}-\frac{1}{k+1})$

Telescoping sum

$=1+\frac{1}{2}$

Simplify

$=\frac{3}{2}$

change order of summation

Telescoping sum

Partial fractions

Telescoping sum

Simplify

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