What is the limit when $n\rightarrow +\infty$ of the following sum?
$$\frac{1}{n^{n-1}}\sum_{k=1}^{n}{k^{n-1}}$$
This sum came up in a probability problem: given $n$ random numbers in $\{1,...,n\}$, what is the expected number of maxima? It turns out that the previous sum is the answer, and it can be derived just applying the definition of expected value with some not too hard calculations.
I was interested in knowing what does that sum approaches (if it approaches any value) when $n$ is large, and plugging big values of $n$ into online sum calculators it seems that the answer is roughly $1.5819$, though i couldn't find a precise answer to what that constant is.
I know that $\sum_{k=0}^{n}{k^m}=\frac{n^{m+1}}{m+1}+\frac{1}{2}n^m+O(n^{m-1})$, but that is for a fixed $m$, while in our case the exponent is not fixed, since it depends on $n$.
I was only able to get, using integrals for estimates, that the sum should be between $1$ and $e$. Does anyone know how to compute the limit (if it actually converges)?
