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Question Number 61804 by maxmathsup by imad last updated on 09/Jun/19

$$calculate\sum _{n=2}^{\mathrm{\infty }}\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k^{n}k!}$$

Commented by maxmathsup by imad last updated on 15/Jun/19

$$S=\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k!}\sum _{n=2}^{\mathrm{\infty }}{\left(\frac{1}{k}\right)}^n=\sum _{k=2}^{\mathrm{\infty }}\frac{w_k}{k!}$$$$w_k=\sum _{n=2}^{\mathrm{\infty }}{\left(\frac{1}{k}\right)}^n=\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{1}{k}\right)}^n-1-\frac{1}{k}=\frac{1}{1-\frac{1}{k}}-1-\frac{1}{k}=\frac{k}{k-1}-1-\frac{1}{k}$$$$=\frac{k-1+1}{k-1}-1-\frac{1}{k}=\frac{1}{k-1}-\frac{1}{k}\Rightarrow S=\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k!}(\frac{1}{k-1}-\frac{1}{k})$$$$=\sum _{k=2}^{\mathrm{\infty }}\frac{1}{(k-1)k!}-\sum _{k=2}^{\mathrm{\infty }}\frac{1}{kk!}$$$$=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k+1)!}-\sum _{k=2}^{\mathrm{\infty }}\frac{1}{kk!}letfind\sum _{k=2}^{\mathrm{\infty }}\frac{1}{kk!}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{kk!}-1$$$$wehave{e}^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^k}{k!}\Rightarrow {\int }_0^x{e}^{t}dt=\sum _{k=0}^{\mathrm{\infty }}{\left[\frac{t^{k+1}}{(k+1)k!}\right]}_0^x+c\Rightarrow$$$$e^x-1=\sum _{k=0}^{\mathrm{\infty }}\frac{x^{k+1}}{(k+1)k!}$$$$x=1\Rightarrow e-1=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(k+1)k!}$$$$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k+1)!}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k(k+1)k!}=\sum _{k=1}^{\mathrm{\infty }}(\frac{1}{k}-\frac{1}{k+1})\frac{1}{k!}=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{kk!}-\sum _{k=1}^{\mathrm{\infty }}\frac{1}{(k+1)k!}$$$$\Rightarrow S=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{kk!}-\sum _{k=1}^{\mathrm{\infty }}\frac{1}{(k+1)k!}-\sum _{k=2}^{\mathrm{\infty }}\frac{1}{kk!}$$$$=1-\{\sum _{k=0}^{\mathrm{\infty }}\frac{1}{(k+1)k!}-1\}=1-\{e-1-1\}=1-e+2=3-e\Rightarrow$$$$S=3-e.$$

Answered by perlman last updated on 09/Jun/19

$$\sum _{n=2}^{\mathrm{\infty }}\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k^{n}k!}=\sum _{k=2}^{+\mathrm{\infty }}\sum _{n=2}^{+\mathrm{\infty }}\frac{1}{k^{n}k!}$$$$justify$$$$\mathrm{\forall }k,n⩾2$$$$\sum _{n=2}^{+\mathrm{\infty }}\sum _{k=2}^{+\mathrm{\infty }}\frac{1}{k!k^n}<\sum _n\frac{e}{2^n}=e$$$$\sum _{n=2}^{\mathrm{\infty }}\sum _{k=2}^{\mathrm{\infty }}\frac{1}{k^{n}k!}=\sum _{k=2}^{+\mathrm{\infty }}\sum _{n=2}^{+\mathrm{\infty }}\frac{1}{k^{n}k!}=\sum _k\frac{1}{k!}\sum _{n=2}\frac{1}{k^n}$$$$=\sum _k\frac{1}{k!}.\frac{1}{k^2}.\frac{k}{k-1}=\sum _{k⩾2}\frac{1}{k!k(k-1)}=\sum _k\frac{k-(k-1)}{k!k(k-1)}=\sum _k\frac{1}{k!(k-1)}-\frac{1}{k!k}$$$$=\sum _k\frac{k-(k-1)}{k!(k-1)}-\sum _k\frac{1}{kk!}$$$$=\sum _k\frac{1}{(k-1)!(k-1)}-\sum _{k=2}^{+\mathrm{\infty }}\frac{1}{k!}-\mathrm{\Sigma }\frac{1}{k!k}$$$$=\sum _{k=2}^{+\mathrm{\infty }}\frac{1}{(k-1)!(k-1)}-\sum _{k=2}^{+\mathrm{\infty }}\frac{1}{k!k}-\sum _{k=0}^{+\mathrm{\infty }}\frac{1}{k!}+\sum _{k=0}^1\frac{1}{k!}$$$$=\frac{1}{(2-1)!(2-1)}-e+1+1=3-e$$$$$$$$$$

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