Find $ \lim _{n\to \infty} \frac{n^1+\dots+n^n}{1^n+\dots+n^n}$ What is the limit of:
$$\lim_{n\to \infty} \frac{n^1+\dots+n^n}{1^n+\dots+n^n} = ?$$
I did some computation with big numbers, I guess it is in the interval $\left(\frac{1}{2},1\right)$.
 A: The answer is $1-1/e$. In fact, one may prove
$$\lim_{n\to\infty}\frac{n^1+n^2+\ldots+n^n}{n^n}=1$$
and
$$\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}=\frac{1}{1-1/e}$$
and the conclusion follows.

The numerator is a geometric series that evaluates to $(n^{n+1}-n)/(n-1)$. One thus have
\begin{align}
\lim_{n\to\infty}\frac{n^1+n^2+\ldots+n^n}{n^n}=\lim_{n\to\infty}\frac{1-n^{-n-1}}{1-1/n}=1
\end{align}
The denominator can be estimated in a quick way, by monotone convergence theorem or, in case this theorem is not available, by the following argument
$$\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}>\lim_{n\to\infty}\sum_{k=0}^K\left(\frac{n-k}n\right)^n=\sum_{k=0}^Ke^{-k}$$
Let $k\to\infty$, we have
$$\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}\ge\frac{1}{1-1/e}$$
On the other hand, since $\ln(1+x)\le x$, we have
\begin{align}
\lim_{n\to\infty}\frac{1^n+2^n+\ldots+n^n}{n^n}&=\lim_{n\to\infty}\sum_{k=0}^{n-1}\left(1-\frac kn\right)^n\\
&=\lim_{n\to\infty}\sum_{k=0}^{n-1}\exp(n\ln(1-k/n))\\
&\le\lim_{n\to\infty}\sum_{k=0}^{n-1}\exp(-k)\\
&=\frac1{1-1/e}
\end{align}
