Find limit $\lim_{n\to \infty}\frac{\sum_{k=1}^n k^n}{n^n}$.

Hint: For fixed $k,$ $$\frac{(n-k)^n}{n^n}=\left(1-\frac{k}n\right)^n\to e^{-k}$$

---

Details:

Note that for $k=0...,n-1$ $$\log\left(1-\frac{k}{n}\right)=-\frac{k}{n}-\frac{k^2}{2n^2}-\cdots\leq -\frac{k}{n}$$

So $$\left(1-\frac{k}{n}\right)^n \leq e^{-k}$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=\begin{cases}0 & k\geq n\\ \left(1-\frac{k}{n}\right)^n& 0\leq k<n\end{cases}$$

Then $|f_n(k)|\leq g(k)=e^{-k}$ and for any $k,$ $\lim_{k\to\infty}f_n(k)\to g(k).$

We have that

$$\frac{\sum\limits_{k=1}^n k^n}{n^n}=\sum_{k=1}^n \left( \frac k n \right)^n=\sum_{k=0}^{n-1} \left( \frac {n-k} n \right)^n=\sum_{k=0}^{n-1} \left( 1-\frac {k} n \right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=\sum_{\substack{k=0 \\ k<n}}^{\infty} \left( 1-\frac {k} n \right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

• $\sqrt[n+1]{\left(1-\frac{k}{n}\right)^n \cdot 1} \leq \frac{n \left(1 - \frac{k}{n}\right) + 1}{n+1} = 1 - \frac{k}{n+1} $

and

• $\left( 1-\frac {k} n \right)^n=e^{n\log \left( 1-\frac {k} n \right)}=e^{n \left( -\frac {k} {n}-\frac {k^2} {2n^2}-\frac {k^3} {3n^3}-\ldots \right)}\le e^{-k}$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

• $\left( 1-\frac {k} n \right)^n\to e^{-k}$

we have

$$\lim_{n\to \infty}\frac{\sum\limits_{k=1}^n k^n}{n^n}=\lim_{n\to \infty} \sum_{k=0}^{n-1} \left( 1-\frac {k} n \right)^n=\lim_{n\to \infty} \sum_{\substack{k=0 \\ k<n}}^{\infty} \left( 1-\frac {k} n \right)^n=\sum_{k=0}^{\infty}e^{-k}=\frac{e}{e-1}$$