I have been trying to solve such sums for a while involving limits of summations but haven't got any luck on this one yet...
$$\lim_{n\to \infty} \sum_{r=0}^n (r/n)^n$$. I first thought of finding a series greater than this one and hoping that would be zero, so that I can claim that the given question is also zero. I could have used integration if there was another n in the denominator. And possibly, I don't think there is any way to sum up the numerator.
On
We can show the limit exists and justify the steps to evaluate using the monotone convergence theorem.
Note that with $k = n - r$ we have
$$\sum_{r=1}^n \left(\frac{r}{n} \right)^n = \sum_{k=0}^{n-1} \left(1-\frac{k}{n} \right)^n = \sum_{k=0}^{\infty} a_{k,n}, $$
where
$$a_{k,n} = \begin{cases}(1 - k/n)^n, \,\,\, k < n \\ 0, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k \geqslant n \end{cases}$$
The sequence $a_{k,n}$ is increasing with respect to $n$ and convergent to $e^{-k}$ as $n \to \infty.$ By the monotone convergence theorem for series it follows that
$$\lim_{n \to \infty}\sum_{r=1}^n \left(\frac{r}{n} \right)^n = \lim_{n \to \infty} \sum_{k=0}^\infty a_{k,n} \\ = \sum_{k=0}^\infty \lim_{n \to \infty}a_{k,n} \\ = \sum_{k=0}^\infty e^{-k} \\ = \frac{1}{1 - e^{-1}}$$
Observe you have \begin{align} \sum^n_{r=1}\left(\frac{r}{n}\right)^n = \sum^n_{r=1}\left(1-\frac{n-r}{n}\right)^n\approx \sum^n_{r=1}e^{-n+r} = e^{-n}\sum^n_{r=1}e^r = e^{-n}\frac{e^{n+1}-e}{e-1} = \frac{e-e^{-n+1}}{e-1}. \end{align} Hence in the limit you get \begin{align} \lim_{n\rightarrow \infty}\sum^n_{r=1}\left(\frac{r}{n}\right)^n=\frac{e}{e-1}. \end{align}