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This question already has an answer here:

How to calculate the following limit?$$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\left(\frac{k}{n}\right)^n$$ It is easy to seem the limit's existence. But I don't know how to calculate its value.

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marked as duplicate by YuiTo Cheng, Leucippus, Theo Bendit, Cesareo, Shailesh Jun 26 at 9:28

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    $\begingroup$ $\lim\limits_{n\to\infty}\left(\dfrac{n-j}{n}\right)^n=e^{-j}$. Looks like $\dfrac{e}{e-1}$? $\endgroup$ – Jonas Meyer Nov 9 '12 at 7:33
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    $\begingroup$ Nice! Why not make it an answer? $\endgroup$ – copper.hat Nov 9 '12 at 7:36
  • $\begingroup$ @Jonas Meyer You are absolutely right! $\endgroup$ – Eastsun Nov 9 '12 at 7:41
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Let $m$ be an arbitrary positive integer. When $n>m$,

$$\sum_{j=0}^{m}\left(1-\frac{j}{n}\right)^n=\sum_{k=n-m}^{n}\left(\frac{k}{n}\right)^n\leq \sum_{k=1}^{n}\left(\frac{k}{n}\right)^n=\sum_{j=0}^{n-1}\left(1-\frac{j}{n}\right)^n\leq\sum_{j=0}^{n-1}e^{-j}<\frac{e}{e-1}.$$

Thus the limit is at most $\dfrac{e}{e-1}$, and taking limits in the inequality

$$\sum_{j=0}^{m}\left(1-\frac{j}{n}\right)^n\leq \sum_{k=1}^{n}\left(\frac{k}{n}\right)^n$$ yields $$\sum_{j=0}^me^{-j}\leq\lim\limits_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^n.$$

Since the right-hand side does not depend on $m$, taking the limit as $m\to\infty$ yields

$$\frac{e}{e-1}\leq \lim\limits_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^n.$$

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