# How to calculate the limit: $\lim_{n\to\infty}\sum_{k=1}^n\big(\frac{k}{n}\big)^n$ [duplicate]

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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.

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

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

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