nidantasnu

2022-07-05

Consider the following sum:
$S\left(n\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{\left(\genfrac{}{}{0}{}{2k+n}{k}\right)}{2k+n}\frac{1}{{2}^{2k}};n=0,1,2,3,...$

Karla Hull

Expert

The sum at hand is a hypergeometric series. Let
${c}_{k}=\frac{1}{n+2k}\left(\genfrac{}{}{0}{}{n+2k}{k}\right)\frac{1}{{2}^{2k}}=\frac{\left(n-1+2k\right)!}{k!\left(n+k\right)!}\frac{1}{{4}^{k}}$
Indeed, the hypergeometric certificate is:
$\frac{{c}_{k+1}}{{c}_{k}}=\frac{1}{4}\frac{\left(n+2k\right)\left(n+2k+1\right)}{\left(n+1+k\right)\left(k+1\right)}$
Meaning that
$\sum _{k=0}^{\mathrm{\infty }}{c}_{k}={c}_{0}\sum _{k=0}^{\mathrm{\infty }}\frac{{\left(n/2\right)}_{k}{\left(n/2+1/2\right)}_{k}}{\left(n+1{\right)}_{k}}\frac{1}{k!}=\frac{1}{n}\cdot {}_{2}{F}_{1}\left(\frac{n}{2},\frac{n+1}{2};n+1;1\right)$
where $\left(a{\right)}_{k}$ denotes Pochhammer symbol. Using Gauss's theorem, applicable for $\mathrm{\Re }\left(c-a-b\right)>0$
${}_{2}{F}_{1}\left(a,b;c;1\right)=\frac{\mathrm{\Gamma }\left(c\right)\mathrm{\Gamma }\left(c-a-b\right)}{\mathrm{\Gamma }\left(c-a\right)\mathrm{\Gamma }\left(c-b\right)}$
we have
$\sum _{k=0}^{\mathrm{\infty }}{c}_{k}=\frac{1}{n}\frac{\mathrm{\Gamma }\left(n+1\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n}{2}+1\right)\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}\stackrel{\text{duplication}}{=}\frac{1}{n}\frac{{2}^{n}{\pi }^{-1/2}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\mathrm{\Gamma }\left(\frac{n}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n}{2}\right)\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}=\frac{{2}^{n}}{n}$
Since $\mathrm{\Gamma }\left(1/2\right)=\sqrt{\pi }$

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