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Question Number 179991 by mnjuly1970 last updated on 05/Nov/22

Answered by mindispower last updated on 05/Nov/22

$$=\sum _{n⩾2}{\int }_0^{\mathrm{\infty }}\frac{1}{m!}{x}^m{e}^{-nx}dx=\frac{1}{m!}{\int }_0^{\mathrm{\infty }}{\left(\frac{y}{n}\right)}^m{e}^{-y}\frac{dy}{n}$$$$=\frac{1}{m!}{\int }_0^{\mathrm{\infty }}\frac{1}{n^{m+1}}{y}^m{e}^{-y}dy=\frac{1}{m!}.\frac{1}{n^{m+1}}{\int }_0^{\mathrm{\infty }}{y}^m{e}^{-y}dy$$$$=\frac{1}{m!}.\frac{1}{n^{m+1}}.\mathrm{\Gamma }(m+1)=\frac{1}{n^{m+1}}$$$$\sum _{m⩾1}\frac{1}{n^{m+1}}=\frac{1}{n^2}.\frac{1}{1-\frac{1}{n}}=\frac{1}{n(n-1)}$$$$\sum _{n⩾2}\frac{1}{n(n-1)}=\mathrm{\Sigma }\frac{1}{n-1}-\frac{1}{n}=1$$$$$$

Commented by mnjuly1970 last updated on 05/Nov/22

$$merceysir$$

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