TITLE: Find $\lim\limits_{n \to \infty}\sum_{k=1}^n\left(\frac{k}{n}\right)^k$.
QUESTION [13 upvotes]: Maybe, we can make use of Tannery's Theorem, or dominated convergence theorem, to exchange the order of the limit and summation:
\begin{align*}
\lim_{n \to \infty}\sum_{k=1}^n\left(\frac{k}{n}\right)^k&=\lim_{n \to \infty}\sum_{k=0}^{n-1}\left[\left(1-\frac{k}{n}\right)^{n}\right]^{\frac{n-k}{n}}=\sum_{k=0}^{\infty}\lim_{n \to \infty}\left[\left(1-\frac{k}{n}\right)^{n}\right]^{\frac{n-k}{n}}=\sum_{k=0}^{\infty} e^{-k}=\frac{e}{e-1}
\end{align*}
This is correct? How to verify that it satisfy the conditons of the theorem?

REPLY [5 votes]: Answer Using Tannery's Theorem and Bernoulli's Inequality
Note that $\left(\frac{n-k}n\right)^{n-k}$ is decreasing in $n$ for $n\gt k$. A proof using Bernoulli's Inequality is given below.
$$
\begin{align}
\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n}\right)^k
&=\lim_{n\to\infty}\sum_{k=0}^{n-1}\left(\frac{n-k}{n}\right)^{n-k}\tag1\\
&=\lim_{n\to\infty}\frac1n+\lim_{n\to\infty}\sum_{k=0}^{n-2}\left(\frac{n-k}{n}\right)^{n-k}\tag2\\
&=0+\sum_{k=0}^\infty e^{-k}\tag3\\[3pt]
&=\frac{e}{e-1}\tag4
\end{align}
$$
Explanation:
$(1)$: substitute $k\mapsto n-k$
$(2)$: isolate the $k=n-1$ term
$(3)$: each term in the sum is no greater than $\left(\frac2{k+2}\right)^2$
$\phantom{(4)\text{:}}$ so we can apply Tannery's Theorem,
$\phantom{(4)\text{:}}$ which is Dominated Convergence for Series $\left(\ell^1\right)$
$(4)$: sum the geometric series

More Detail on Step $\boldsymbol{(3)}$
Step $(3)$ is a bit tricky. We isolate the $k=n-1$ term so that the terms in the remaining sum are no greater than $\left(\frac2{k+2}\right)^2$. For $n\lt k+2$, the terms are $0$ (or missing). For $n=k+2$, the term is
$$
\left(\frac{n-k}{n}\right)^{n-k}=\left(\frac2{k+2}\right)^2\tag5
$$
For $n\ge k+2$, $\left(\frac{n-k}{n}\right)^{n-k}$ decreases, as shown below, from $\left(\frac2{k+2}\right)^2$ to $e^{-k}$.
We can then apply Tannery's Theorem because $\sum\limits_{k=0}^\infty\left(\frac2{k+2}\right)^2\lt\infty$.

Bernoulli says $\boldsymbol{\left(\frac{n-k}n\right)^{n-k}}$ is Decreasing in $\boldsymbol{n}$
$$
\begin{align}
\frac{\left(\frac{n-k}n\right)^{n-k}}{\left(\frac{n-k+1}{n+1}\right)^{n-k+1}}
&=\frac{n}{n-k}\left(\frac{n-k}n\frac{n+1}{n-k+1}\right)^{n-k+1}\tag6\\
&=\frac{n}{n-k}\left(1-\frac{k}{(n-k+1)n}\right)^{n-k+1}\tag7\\[3pt]
&\ge\frac{n}{n-k}\left(1-\frac{k}{n}\right)\tag8\\[9pt]
&=1\tag9
\end{align}
$$
Explanation:
$(6)$: algebra
$(7)$: algebra
$(8)$: Bernoulli's Inequality
$(9)$: algebra