lim-n-1-n-2-r-1-n-r-e-r-n-

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

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

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

A_{n} = \frac{1}{n^{2}} \sum_{k = 1}^{n} k e^{\frac{k}{n}} \Rightarrow A_{n} = \frac{1}{n} \sum_{k = 1}^{n} \frac{k}{n} e^{\frac{k}{n}} so A_{n} is a Rieman sum and

\lim_{n \to + \infty} A_{n} = \int_{0}^{1} {xe}^{x} dx = {[x e^{x}]}_{0}^{1} - \int_{0}^{1} e^{x} dx

= e - (e - 1) = 1

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

\frac{1}{n} \lim_{n \to \infty} \underset{r = 1}{\sum^{n}} \frac{r}{n} e^{\frac{r}{n}} = \int_{0}^{1} {x e}^{x} d x = {[{x e}^{x}]}_{0}^{1} - {[e^{x}]}_{0}^{1} = 1

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

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