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.
How to calculate the limit: $\lim_{n\to\infty}\sum_{k=1}^n\big(\frac{k}{n}\big)^n$
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limits
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0@Jonas Meye$r$ You are absolutely right! – 2012-11-09
1 Answers
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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
$\frac{e}{e-1}\leq \lim\limits_{n\to\infty}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^n.$