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Question Number 35035 by abdo mathsup 649 cc last updated on 14/May/18

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

Answered by tanmay.chaudhury50@gmail.com last updated on 14/May/18

$$T_k=\frac{k}{(k+1)!}$$$$=\frac{k+1-1}{(k+1)!}$$$$T_k=\frac{1}{k!}-\frac{1}{(k+1)!}$$$$T_1=\frac{1}{1!}-\frac{1}{2!}$$$$T_2=\frac{1}{2!}-\frac{1}{3!}$$$$T_3=\frac{1}{3!}-\frac{1}{4!}$$$$................$$$$...............$$$$T_n=\frac{1}{n!}-\frac{1}{(n+1)!}$$$$so{T}_1+T_2+T_3+....+T_n=\frac{1}{1!}-\frac{1}{(n+1)!}$$$$=1-\frac{1}{(n+1)!}$$

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