siutulysr5

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

Frauental91

Expert

Let $l=n-k$, we have
$\sum _{n=4}^{\mathrm{\infty }}\sum _{k=2}^{n-2}{\left(\begin{array}{c}n\\ k\end{array}\right)}^{-1}=\sum _{k=2}^{\mathrm{\infty }}\sum _{n=k+2}^{\mathrm{\infty }}{\left(\begin{array}{c}n\\ k\end{array}\right)}^{-1}$
$=\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}{\left(\begin{array}{c}k+l\\ k\end{array}\right)}^{-1}=\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}\frac{k!l!}{\left(k+l\right)!}$
$=\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}\left(k+l+1\right)\frac{\mathrm{\Gamma }\left(k+1\right)\mathrm{\Gamma }\left(l+1\right)}{\mathrm{\Gamma }\left(k+l+2\right)}$
$=\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}\left(k+l+1\right){\int }_{0}^{1}{t}^{k}{\left(1-t\right)}^{l}dt$
$={\int }_{0}^{1}\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}\left[\left(k+l+1\right){t}^{k}{\left(1-t\right)}^{l}\right]dt$
Notice when s and t are independent, we have
$\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}\left(k+l+1\right){t}^{k}{s}^{l}=\sum _{k=2}^{\mathrm{\infty }}\sum _{l=2}^{\mathrm{\infty }}\left(t\frac{d}{dt}+s\frac{d}{ds}+1\right){t}^{k}{s}^{l}$
$=\left(t\frac{d}{dt}+s\frac{d}{ds}+1\right)\frac{{t}^{2}{s}^{2}}{\left(1-t\right)\left(1-s\right)}$

Aaron Hughes

Expert

Partial Fractions and Telescoping Sums
change order of summation
$\sum _{n=4}^{\mathrm{\infty }}\sum _{k=2}^{n-2}\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}=\sum _{k=2}^{\mathrm{\infty }}\sum _{n=k+2}^{\mathrm{\infty }}\frac{1}{\left(\begin{array}{c}n\\ k\end{array}\right)}$
$=\sum _{k=2}^{\mathrm{\infty }}\frac{k}{k-1}\sum _{n=k+2}^{\mathrm{\infty }}\left(\frac{1}{\left(\begin{array}{c}n-1\\ k-1\end{array}\right)}-\frac{1}{\left(\begin{array}{c}n\\ k-1\end{array}\right)}\right)$
Telescoping sum
$=\sum _{k=2}^{\mathrm{\infty }}\frac{k}{k-1}\frac{1}{\left(\begin{array}{c}k+1\\ k-1\end{array}\right)}$
$=\sum _{k=2}^{\mathrm{\infty }}\frac{2}{\left(k-1\right)\left(k+1\right)}$
Partial fractions $\sum _{k=2}^{\mathrm{\infty }}\left(\frac{1}{k-1}-\frac{1}{k+1}\right)$
Telescoping sum
$=1+\frac{1}{2}$
Simplify
$=\frac{3}{2}$

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