Just as the topic ask how to evaluate

$\sum _{k=1}^{\mathrm{\infty}}\frac{(18)[(k-1)!{]}^{2}}{(2k)!}.$

$\sum _{k=1}^{\mathrm{\infty}}\frac{(18)[(k-1)!{]}^{2}}{(2k)!}.$

Boilanubjaini8f
2022-06-16
Answered

Just as the topic ask how to evaluate

$\sum _{k=1}^{\mathrm{\infty}}\frac{(18)[(k-1)!{]}^{2}}{(2k)!}.$

$\sum _{k=1}^{\mathrm{\infty}}\frac{(18)[(k-1)!{]}^{2}}{(2k)!}.$

You can still ask an expert for help

gaiageoucm5p

Answered 2022-06-17
Author has **20** answers

Notice that

$\frac{(k-1){!}^{2}}{(2k)!}=\frac{\mathrm{\Gamma}(k)\mathrm{\Gamma}(k)}{\mathrm{\Gamma}(2k+1)}=\frac{1}{2k}B(k,k)=\frac{1}{2k}{\int}_{0}^{1}{x}^{k-1}(1-x{)}^{k-1}\mathrm{d}x$

Thus

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

$\frac{(k-1){!}^{2}}{(2k)!}=\frac{\mathrm{\Gamma}(k)\mathrm{\Gamma}(k)}{\mathrm{\Gamma}(2k+1)}=\frac{1}{2k}B(k,k)=\frac{1}{2k}{\int}_{0}^{1}{x}^{k-1}(1-x{)}^{k-1}\mathrm{d}x$

Thus

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

mravinjakag

Answered 2022-06-18
Author has **4** answers

In order to fill in on the last equality, define

$f(t)=-2{\int}_{0}^{1}\frac{\mathrm{ln}(1-t(1-{u}^{2}))}{1-{u}^{2}}\mathrm{d}u$

Clearly $f(0)=0$, and we are interested in computing $f\left(\frac{1}{4}\right)$

${f}^{\mathrm{\prime}}(t)=2{\int}_{0}^{1}\frac{\mathrm{d}u}{1+t(1-{u}^{2})}\stackrel{u=\sqrt{\frac{1-t}{t}}\mathrm{tan}(\varphi )}{=}{\int}_{0}^{\mathrm{arcsin}(\sqrt{t})}\frac{2\mathrm{d}\varphi}{\sqrt{t(1-t)}}=\frac{2\mathrm{arcsin}(\sqrt{t})}{\sqrt{t(1-t)}}=\phantom{\rule{0ex}{0ex}}2\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{arcsin}}^{2}(\sqrt{t})$

Thus

$f\left(\frac{1}{4}\right)={\int}_{0}^{\frac{1}{4}}2\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{arcsin}}^{2}(\sqrt{t})=2{\mathrm{arcsin}}^{2}\left(\frac{1}{2}\right)=\frac{{\pi}^{2}}{18}$

$f(t)=-2{\int}_{0}^{1}\frac{\mathrm{ln}(1-t(1-{u}^{2}))}{1-{u}^{2}}\mathrm{d}u$

Clearly $f(0)=0$, and we are interested in computing $f\left(\frac{1}{4}\right)$

${f}^{\mathrm{\prime}}(t)=2{\int}_{0}^{1}\frac{\mathrm{d}u}{1+t(1-{u}^{2})}\stackrel{u=\sqrt{\frac{1-t}{t}}\mathrm{tan}(\varphi )}{=}{\int}_{0}^{\mathrm{arcsin}(\sqrt{t})}\frac{2\mathrm{d}\varphi}{\sqrt{t(1-t)}}=\frac{2\mathrm{arcsin}(\sqrt{t})}{\sqrt{t(1-t)}}=\phantom{\rule{0ex}{0ex}}2\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{arcsin}}^{2}(\sqrt{t})$

Thus

$f\left(\frac{1}{4}\right)={\int}_{0}^{\frac{1}{4}}2\frac{\mathrm{d}}{\mathrm{d}t}{\mathrm{arcsin}}^{2}(\sqrt{t})=2{\mathrm{arcsin}}^{2}\left(\frac{1}{2}\right)=\frac{{\pi}^{2}}{18}$

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