Boilanubjaini8f

2022-06-16

Just as the topic ask how to evaluate
$\sum _{k=1}^{\mathrm{\infty }}\frac{\left(18\right)\left[\left(k-1\right)!{\right]}^{2}}{\left(2k\right)!}.$

gaiageoucm5p

Notice that
$\frac{\left(k-1\right){!}^{2}}{\left(2k\right)!}=\frac{\mathrm{\Gamma }\left(k\right)\mathrm{\Gamma }\left(k\right)}{\mathrm{\Gamma }\left(2k+1\right)}=\frac{1}{2k}B\left(k,k\right)=\frac{1}{2k}{\int }_{0}^{1}{x}^{k-1}\left(1-x{\right)}^{k-1}\mathrm{d}x$
Thus
$\begin{array}{rcl}\sum _{k=1}^{\mathrm{\infty }}\frac{\left(k-1\right){!}^{2}}{\left(2k\right)!}& =& {\int }_{0}^{1}\left(\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2k}{x}^{k-1}\left(1-x{\right)}^{k-1}\right)\mathrm{d}x={\int }_{0}^{1}\frac{-\mathrm{ln}\left(1-x+{x}^{2}\right)}{2x\left(1-x\right)}\mathrm{d}x\\ & =& -2{\int }_{-1/2}^{1/2}\frac{\mathrm{ln}\left(3/4+{u}^{2}\right)}{1-4{u}^{2}}\mathrm{d}u=-2{\int }_{0}^{1}\frac{\mathrm{ln}\left(\left(3+{u}^{2}\right)/4\right)}{1-{u}^{2}}\mathrm{d}u=\frac{{\pi }^{2}}{18}\end{array}$

mravinjakag

In order to fill in on the last equality, define
$f\left(t\right)=-2{\int }_{0}^{1}\frac{\mathrm{ln}\left(1-t\left(1-{u}^{2}\right)\right)}{1-{u}^{2}}\mathrm{d}u$
Clearly $f\left(0\right)=0$, and we are interested in computing $f\left(\frac{1}{4}\right)$
${f}^{\mathrm{\prime }}\left(t\right)=2{\int }_{0}^{1}\frac{\mathrm{d}u}{1+t\left(1-{u}^{2}\right)}\stackrel{u=\sqrt{\frac{1-t}{t}}\mathrm{tan}\left(\varphi \right)}{=}{\int }_{0}^{\mathrm{arcsin}\left(\sqrt{t}\right)}\frac{2\mathrm{d}\varphi }{\sqrt{t\left(1-t\right)}}=\frac{2\mathrm{arcsin}\left(\sqrt{t}\right)}{\sqrt{t\left(1-t\right)}}=\phantom{\rule{0ex}{0ex}}2\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{arcsin}}^{2}\left(\sqrt{t}\right)$
Thus
$f\left(\frac{1}{4}\right)={\int }_{0}^{\frac{1}{4}}2\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{arcsin}}^{2}\left(\sqrt{t}\right)=2{\mathrm{arcsin}}^{2}\left(\frac{1}{2}\right)=\frac{{\pi }^{2}}{18}$

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