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Question Number 8553 by Sopheak last updated on 16/Oct/16

Commented by Yozzias last updated on 16/Oct/16

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

Answered by Yozzias last updated on 16/Oct/16

$$\mathrm{S}=\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\frac{\mathrm{r}}{(\mathrm{r}+1)!}=\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}[\frac{1}{\mathrm{r}!}-\frac{1}{(\mathrm{r}+1)!}]=\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}\{\mathrm{f}\left(\mathrm{r}\right)-\mathrm{f}(\mathrm{r}+1)\}$$$$\mathrm{S}=\{\mathrm{f}\left(1\right)-\mathrm{f}\left(2\right)\}+\{\mathrm{f}\left(2\right)-\mathrm{f}\left(3\right)\}+\{\mathrm{f}\left(3\right)-\mathrm{f}\left(4\right)\}$$$$+\{\mathrm{f}\left(4\right)-\mathrm{f}\left(5\right)\}+...+\{\mathrm{f}(\mathrm{n}-1)-\mathrm{f}\left(\mathrm{n}\right)\}+\{\mathrm{f}\left(\mathrm{n}\right)-\mathrm{f}(\mathrm{n}+1)\}$$$$\mathrm{S}=\mathrm{f}\left(1\right)-\mathrm{f}(\mathrm{n}+1)$$$$\mathrm{S}=\frac{1}{1!}-\frac{1}{(\mathrm{n}+1)!}=1-\frac{1}{(\mathrm{n}+1)!}$$$$\mathrm{Corollary}:\underset{\mathrm{n}\to \mathrm{\infty }}{\lim S}=1-0=1$$

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