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

  • $\begingroup$ Are you familiar with Riemann sums associated to Riemann integrals? $\endgroup$ – Did Aug 26 '12 at 12:44
  • $\begingroup$ There is a celebre limit that states that if $\lim_{n\to\infty} a_{n} = a \geq 0$ and $\lim_{n\to\infty} b_{n} = b \geq 0$ then we have that $ \lim_{n\to\infty} \frac{a_{1} b_{n} + a_{2} b_{n-1}+ \cdots +a_{n} b_{1}}{n} =ab.$ I think this could be an alternative. $\endgroup$ – user 1357113 Aug 26 '12 at 15:44

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

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


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

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

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

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


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

(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.

  • $\begingroup$ @RagibZaman: I understood what you are pointing out. Since $x \mapsto x^M$ is increasing, $$ \frac{1}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^M \leq \int_{0}^{1} x^M \; dx,$$ which makes the argument much simpler. $\endgroup$ – Sangchul Lee Aug 26 '12 at 13:11
  • $\begingroup$ Actually, I misunderstood the subtlety of the argument initially, which is why I deleted my comment. It was not as simple as I had thought since the right hand bound doesn't tend to $0$ with $N$ immediately, but rather it depends on $M$ as well. Luckily, you've turned my ignorant comment into another nice finish to you proof! $\endgroup$ – Ragib Zaman Aug 26 '12 at 13:13
  • $\begingroup$ Again a nice & simple proof! (+1) $\endgroup$ – user 1357113 Aug 26 '12 at 14:28

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 $0.$

  • $\begingroup$ +1 for the nice estimation of a rapidly decreasing bound! $\endgroup$ – Sangchul Lee Aug 26 '12 at 13:08
  • $\begingroup$ Nice answer. (+1) $\endgroup$ – user 1357113 Aug 26 '12 at 14:28

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