TITLE: How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$?
QUESTION [26 upvotes]: I can show that the following limit exists but
I am having difficulties to find it. It is
$$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}$$
Can someone please help me?

REPLY [19 votes]: Just for reference: With aid of some fancy theorem, you can skip most of hard analysis. As in other answers, we begin by writing
$$ \sum_{k=1}^{n} \left( \frac{k}{n}\right)^n
\ \overset{k \to n-k}{=} \ \sum_{k=0}^{n-1} \left( 1 - \frac{k}{n}\right)^n
\ = \ \sum_{k=0}^{\infty} \left( 1 - \frac{k}{n}\right)^n \mathbf{1}_{\{k < n\}}, $$
where $\mathbf{1}_{\{k < n\}}$ is the indicator function which takes value $1$ if $k < n$ and $0$ otherwise. Now for each $0 \leq k < n$, utilizing the inequality $\log(1-x) \leq -x$ which holds for all $x \in [0,1)$ shows that
$$ \left( 1 - \frac{k}{n}\right)^n
= e^{n \log(1 - \frac{k}{n})}
\leq e^{-k}. $$
Since $\sum_{k=0}^{\infty} e^{-k} < \infty$, by the dominated convergence theorem we can interchange the infinite sum and the limit:
$$ \lim_{n\to\infty} \sum_{k=1}^{n} \left( \frac{k}{n}\right)^n
= \sum_{k=0}^{\infty} \lim_{n\to\infty} \left( 1 - \frac{k}{n}\right)^n \mathbf{1}_{\{k < n\}}
= \sum_{k=0}^{\infty} e^{-k}
= \frac{1}{1 - e^{-1}}. $$