Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

Question Number 83629 by M±th+et£s last updated on 04/Mar/20

$$showthat$$$$\underset{n,k=0}{\sum ^{\mathrm{\infty }}}\frac{n!k!}{(n+k+2)!}=\frac{{\pi }^2}{6}$$

Answered by Kamel Kamel last updated on 04/Mar/20

$$S\left(x\right)=\underset{n=0}{\sum ^{+\mathrm{\infty }}}\underset{k=0}{\sum ^{+\mathrm{\infty }}}\frac{n!k!}{(n+k+2)!}{x}^{n+k+2}=\underset{n=0}{\sum ^{+\mathrm{\infty }}}\underset{k=0}{\sum ^{+\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(k+1)}{\mathrm{\Gamma }(n+k+2)(n+k+2)}{x}^{n+k+2}$$$$S^{\prime }\left(x\right)=\underset{n=0}{\sum ^{+\mathrm{\infty }}}\underset{k=0}{\sum ^{+\mathrm{\infty }}}{x}^{n+k+1}{\int }_0^1{t}^n{(1-t)}^{k}dt$$$$=x{\int }_0^1\frac{1}{1-xt}\frac{1}{1-x(1-t)}dt$$$$\frac{1}{(1-xt)(1-x+xt)}=\frac{1}{2-x}(\frac{1}{1-xt}+\frac{1}{1-x+xt})$$$$\therefore S^{\prime }\left(x\right)=\frac{1}{2-x}{\int }_0^1(\frac{1}{1-xt}+\frac{1}{1-x+xt})xdt=-\frac{2}{2-x}Ln(1-x)$$$$=-\frac{2}{2-x}Ln(1-x)$$$$\therefore S=-2{\int }_0^1\frac{Ln(1-x)}{2-x}dx\stackrel{t=2-x}{=}-2{\int }_1^2\frac{Ln(t-1)}{t}dt$$$$\stackrel{z=\frac{1}{t}}{=}-2{\int }_{\frac{1}{2}}^1\frac{Ln(1-z)-Ln\left(z\right)}{z}dz$$$$=2({Li}_2\left(1\right)-{Li}_2\left(\frac{1}{2}\right))-{Ln}^2\left(2\right)$$$${Li}_2(1-x)+{Li}_2\left(x\right)=\frac{{\pi }^2}{6}-Ln\left(x\right)Ln(1-x)$$$$\therefore {Li}_2\left(\frac{1}{2}\right)=\frac{{\pi }^2}{12}-\frac{1}{2}{Ln}^2\left(2\right)$$$$So:S=2(\frac{{\pi }^2}{6}-\frac{{\pi }^2}{12}+\frac{1}{2}{Ln}^2\left(2\right))-{Ln}^2\left(2\right)=\frac{{\pi }^2}{6}$$

Commented by M±th+et£s last updated on 04/Mar/20

$$nicesolutionsirthankyou$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com