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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.

Linus Rastegar
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Vedant Parwal
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2 Answers2

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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}

Jacky Chong
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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

$$\lim_{n \to \infty}\sum_{r=1}^n \left(\frac{r}{n} \right)^n = \lim_{n \to \infty} \sum_{k=0}^\infty a_{k,n} \\ = \sum_{k=0}^\infty \lim_{n \to \infty}a_{k,n} \\ = \sum_{k=0}^\infty e^{-k} \\ = \frac{1}{1 - e^{-1}}$$

RRL
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