Question Number 64268 by Chi Mes Try last updated on 16/Jul/19 | ||
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}^{\mathrm{99}} +\mathrm{2}^{\mathrm{99}} +...+\:{n}^{\mathrm{99}} }{{n}^{\mathrm{100}} }\:= \\ $$ | ||
Commented by Prithwish sen last updated on 16/Jul/19 | ||
$$\mathrm{lim}_{\mathrm{n}−\rightarrow\infty} \frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{r}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\frac{\mathrm{r}}{\mathrm{n}}\right)^{\mathrm{99}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{99}} \mathrm{dx} \\ $$$$=\left[\:\frac{\mathrm{x}^{\mathrm{100}} }{\mathrm{100}}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{100}} \\ $$ | ||