Kasey Haley

2022-01-22

How to find the sum of this series?
$\sum _{k=0}^{\mathrm{\infty }}\frac{{2}^{k}}{\left(\begin{array}{c}2k+1\\ k\end{array}\right)}$

trasahed

Expert

We have, using the Euler Beta function:
$\sum _{k=0}^{+\mathrm{\infty }}\frac{{2}^{k}}{\left(\begin{array}{c}2k+1\\ k\end{array}\right)}=\sum _{k=0}^{+\mathrm{\infty }}\frac{{2}^{k}\mathrm{\Gamma }\left(k+1\right)\mathrm{\Gamma }\left(k+2\right)}{\mathrm{\Gamma }\left(2k+2\right)}$
$=\sum _{k=0}^{+\mathrm{\infty }}{2}^{k}\left(k+1\right){\int }_{0}^{1}{x}^{k}{\left(1-x\right)}^{k}dx$
$={\int }_{0}^{1}\frac{dx}{{\left(1-2x\left(1-x\right)\right)}^{2}}=\frac{\pi }{2}+1$

trovabile4p

Expert

Rewrite binomial coefficient with factorials
$\sum _{k=0}^{\mathrm{\infty }}\frac{{2}^{k}}{\left(\begin{array}{c}2k+1\\ k\end{array}\right)}=\sum _{k=0}^{\mathrm{\infty }}{2}^{k}\frac{k!\left(k+1\right)!}{\left(2k+1\right)!}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(k+1\right)!}{\left(2k+1\right)!}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{k!}{\left(2k+1\right)!!}+\sum _{k=0}^{\mathrm{\infty }}\frac{kk!}{\left(2k+1\right)!!}$
$=\sum _{k=0}^{\mathrm{\infty }}\frac{k!}{\left(2k+1\right)}+\sum _{k=0}^{\mathrm{\infty }}\left(\frac{k!}{\left(2k-1\right)!!}-\frac{\left(k+1\right)!}{\left(2k+1\right)!!}\right)$
$=\frac{\pi }{2}+1$

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