Tricky limit of summation

Question

Find the limit as $n$ tends to infinity of

$$\frac{e^{1/n}}{n^2}+2\frac{e^{2/n}}{n^2}+3\frac{e^{3/n}}{n^2} + \ldots + n \frac{e}{n^2}$$

Any help would be thoroughly appreciated.

@ajotatxe yes it is $n\frac{e^{n/n}}{n^2}=\frac{e}{n}$ – Jimmy R. Oct 23 '14 at 16:49 — Jimmy R., Oct 23 '14 at 16:49

Michael Hardy · Answer 1 · 2014-10-28T18:38:32.440

10

It's a Riemann sum that approaches $$ \int_0^1 xe^x\,dx. $$ The partition is $0,\ 1/n,\ 2/n,\ 3/n,\ \ldots,\ n/n$, so $\Delta x=1/n$ in every subinterval.

Note that $\displaystyle\int x\Big( e^x\,dx\Big) = \int x\,dv = xv-\int v\,dx$ etc.

edited Oct 28 '14 at 18:38

answered Oct 23 '14 at 16:50

Michael Hardy

1

score 0 · Answer 2 · answered Oct 23 '14 at 18:16

0

The sum can be expressed on a closed form. Then, the asymptotic expansion leads to the limit equal to 1.

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answered Oct 23 '14 at 18:16

JJacquelin

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Tricky limit of summation

2 Answers2