How to solve this limit related to series?

Question

How to solve the following limit? $$\lim_{N\rightarrow+\infty}\frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^N$$

Answer 1

Fix a positive integer $M$. Then for $N \geq M$, we have

$$ 0 \leq \frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^N \leq \frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^M. $$

Thus

$$ \begin{align*}0 &\leq \liminf_{N\to\infty} \frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^N \leq \limsup_{N\to\infty} \frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^N \\ &\leq \lim_{N\to\infty} \frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^M = \int_{0}^{1} x^M \; dx = \frac{1}{M+1}. \end{align*}$$

Now taking $M \to \infty$, both liminf and limsup vanishes. Therefore the limit converges to 0.

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Here is a much simpler argument: Since $x \mapsto x^N$ is increasing for $x \geq 0$, we have

$$ \left(\frac{k}{N}\right)^{N}\frac{1}{N} \leq \int_{\frac{k}{N}}^{\frac{k+1}{N}} x^{N} \; dx. $$

Thus

$$ 0 \leq \frac{1}{N}\sum_{k=0}^{N-1}\left(\frac{k}{N}\right)^N \leq \sum_{k=0}^{N-1}\int_{\frac{k}{N}}^{\frac{k+1}{N}} x^{N} \; dx = \int_{0}^{1}x^N \; dx = \frac{1}{N+1}.$$

(Here the beginning point of the summation index is changed to $k = 0$ instead of $k = 1$, which makes no difference.) Thus taking $N \to \infty$ proves the vanishing of the limit.

Answer 2

Fix some large $M$ and suppose $N>M.$ Split the series into the first $N-M-1$ and the last $M$ terms like this: $$ \frac{1}{N} \sum_{k=1}^{N-1} \left( \frac{k}{N} \right)^N =\frac{1}{N} \sum_{k=1}^{N-M-1} \left( \frac{k}{N} \right)^N +\frac{1}{N} \sum_{k=N-M}^{N-1} \left( \frac{k}{N} \right)^N .$$

The summand of the first sum is maximized for $k=N-M-1$ so

$$\frac{1}{N} \sum_{k=1}^{N-M-1} \left( \frac{k}{N} \right)^N< \frac{N-M-1}{N} \left( \frac{N-M-1}{N} \right)^N< \left( \frac{N-M-1}{N} \right)^N \to \exp(-M-1)$$

as $N\to \infty.$

The summand of the second sum is maximized for $k=N-1$ so $$\frac{1}{N} \sum_{k=N-M}^{N-1} \left( \frac{k}{N} \right)^N< \frac{M}{N} \left(\frac{N-1}{N} \right)^N < \frac{M}{N} \to 0$$

as $N\to\infty.$

Thus for sufficiently large $N$, we have $$ \frac{1}{N} \sum_{k=1}^{N-1} \left( \frac{k}{N} \right)^N < \exp(-M-1).$$

For any $\epsilon>0$ we could have chosen $M$ large enough so that $\exp(-M-1) < \epsilon,$ so the limit is $0.$