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Question Number 124738 by mnjuly1970 last updated on 05/Dec/20

$$...nicecalculus...$$$$simplelimit::$$$${lim}_{n\to \mathrm{\infty }}\left\{\frac{1^{a+1}+2^{a+1}+...+n^{a+1}}{n(1^a+2^a+....n^a)}\right\}=?$$$$wherea\ne -2,-1$$

Commented by PRITHWISH SEN 2 last updated on 07/Dec/20

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{\underset{\mathrm{r}=1}{\sum ^{\mathrm{n}}}{\left(\frac{\mathrm{r}}{\mathrm{n}}\right)}^{\mathrm{a}+1}}{\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}^{\mathrm{a}}}=\frac{{\int }_0^1{\mathrm{x}}^{\mathrm{a}+1}\mathrm{dx}}{{\int }_0^1{\mathrm{x}}^{\mathrm{a}}\mathrm{dx}}=\frac{\mathrm{a}+1}{\mathrm{a}+2}$$

Commented by mnjuly1970 last updated on 14/Dec/20

$$thankyousir..$$

Commented by PRITHWISH SEN 2 last updated on 14/Dec/20

$$\mathrm{welcome}$$

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