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Question Number 64444 by mathmax by abdo last updated on 18/Jul/19

$$calculate{A}_n=\sum _{k=0}^n\frac{k}{(k+1)!}$$

Commented by Prithwish sen last updated on 18/Jul/19

$$\frac{1}{\mathrm{k}!}-\frac{1}{(\mathrm{k}+1)!}=1$$

Commented by mathmax by abdo last updated on 22/Jul/19

$$A_n=\sum _{k=0}^n\frac{k+1-1}{(k+1)!}=\sum _{k=0}^n(\frac{1}{k!}-\frac{1}{(k+1)!})let{u}_n=\frac{1}{n!}\Rightarrow$$$$A_n=\sum _{k=0}^n(u_k-u_{k+1})=u_0-u_1+u_1-u_2+....+u_n-u_{n+1}$$$$=u_0-u_{n+1}=1-\frac{1}{(n+1)!}$$$$so{A}_n=1-\frac{1}{(n+1)!}.$$

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