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Question Number 101582 by Rohit@Thakur last updated on 03/Jul/20

$$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n^2}\underset{r=1}{\sum ^n}r{e}^{r/n}=$$

Answered by mathmax by abdo last updated on 03/Jul/20

$${\mathrm{A}}_{\mathrm{n}}=\frac{1}{{\mathrm{n}}^2}\sum _{\mathrm{k}=1}^{\mathrm{n}}\mathrm{k}{\mathrm{e}}^{\frac{\mathrm{k}}{\mathrm{n}}}\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{\mathrm{k}}{\mathrm{n}}{\mathrm{e}}^{\frac{\mathrm{k}}{\mathrm{n}}}\mathrm{so}{\mathrm{A}}_{\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{Rieman}\mathrm{sum}\mathrm{and}$$$${\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\mathrm{xe}}^{\mathrm{x}}\mathrm{dx}={\left[\mathrm{x}{\mathrm{e}}^{\mathrm{x}}\right]}_0^1-{\int }_0^1{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}$$$$=\mathrm{e}-(\mathrm{e}-1)=1$$

Answered by Dwaipayan Shikari last updated on 03/Jul/20

$$\frac{1}{n}\underset{n\to \mathrm{\infty }}{\lim }\underset{r=1}{\sum ^n}\frac{r}{n}{e}^{\frac{r}{n}}={\int }_0^1{xe}^{x}dx={\left[{xe}^x\right]}_0^1-{\left[e^x\right]}_0^1=1$$

Commented by Rohit@Thakur last updated on 04/Jul/20

$$Thankyousir$$

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