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Question Number 81768 by Power last updated on 15/Feb/20

Commented by mathmax by abdo last updated on 15/Feb/20

$$S_n=\sum _{k=1}^n\frac{k}{(k+1)!}=\sum _{k=1}^n\frac{k+1-1}{(k+1)!}=\sum _{k=1}^n(\frac{1}{k!}-\frac{1}{(k+1)!})$$$$=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{n!}-\frac{1}{(n+1)!}=1-\frac{1}{(n+1)!}$$

Commented by mr W last updated on 15/Feb/20

$$answerseeQ81506!$$

Commented by Power last updated on 15/Feb/20

$$\mathrm{thank}\mathrm{you}$$

Commented by abdomathmax last updated on 15/Feb/20

$$youarewelcome$$

Answered by mr W last updated on 15/Feb/20

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

Commented by mr W last updated on 15/Feb/20

$$sir:inQ81506igaveyouallthedetails$$$$fortheproof.$$

Commented by Power last updated on 15/Feb/20

$$\mathrm{sir}\mathrm{prove}\mathrm{thst}\mathrm{pls}$$

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