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How does one prove the following limit without integration? $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\Bigl(\frac{i}{n}\Bigr)^k=\frac{1}{k+1}$$ I was doing some calculus in my free time and tried to prove $\int_{0}^{x}x^kdx=\frac{x^{k+1}}{k+1}$ using the Riemann Sum but got stumped at this infinite sum.

Edit: I fixed the error in my limit and would like an answer that could be understood by a third year student but if that isn't possible then I am willing to spend some time learning these functions in the current answers.

Alex Maslach
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    You probably committed a typo somewhere. $\frac{1}{k+1}<1$ for $k>0$. But LHS summation clearly contains $1$, thus $\geq1$. Was it supposed to be $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\Bigl(\frac{i}{n}\Bigr)^k=\frac{1}{k+1}$$? – rtybase Sep 08 '19 at 07:24
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    If so, then you can use this form of Reimann sums. – rtybase Sep 08 '19 at 07:28
  • @Alex Maslach : Do you want an answer which is understood by someone how is a beginner to calculus or is it ok to give an answer uisng any level of mathematics? Reason for asking is because in Claude answer below which is mathematically correct, it very likely that a beginner is not familiar with zeta function, asymptotic notation and growth rate of the harmonic numbers etc. – Nilotpal Sinha Sep 08 '19 at 08:03
  • @Nilos the answer Claude provided is for $\frac{S_n}{n}=\frac{1}{n}\sum\limits_{i=1}^{n}\Bigl(\frac{i}{n}\Bigr)^k$ which is not indicated in the question (thus my question to OP) and we (there is an army of people willing to answer this question, I can tell :) ) all are waiting for OP to fix the typo. – rtybase Sep 08 '19 at 08:07
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    @rtybase Even if the OP fixes the typpo I don't think some of the ingredients used in Calude's answer will be known to a beginner so I asked OP what what level of mathematics is he expecting an answer :) – Nilotpal Sinha Sep 08 '19 at 08:10
  • @Nilos I always love answers exceeding my knowledge, it drives me to look for more details. And I always consider MSE as a system not just for doing homeworks. – rtybase Sep 08 '19 at 08:13
  • Perhaps Stolz–Cesàro theorem? – kingW3 Sep 08 '19 at 19:00

4 Answers4

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There is a typo in your limit, the correct version should be $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n \left(\frac{i} {n}\right)^k = \frac{1}{k+1}$$

When $k$ is a positive integer, there is an elementary proof of this without using any concept of Riemann sums.

For positive integers $i$ and $k$, we have

$$\begin{align}i^k \le \prod_{\ell=0}^{k-1} (i+\ell) &= \frac{(i+k) - (i-1)}{k+1} )\prod_{\ell=0}^{k-1} (i+\ell)\\ &= \frac{1}{k+1} \left[\prod_{\ell=0}^{k}(i+\ell) - \prod_{\ell=0}^{k}(i - 1 + \ell)\right]\end{align} $$ The sums for finite $n$ is bounded from above by a telescoping sum.
Evaluating the telescoping sum give us: $$\frac1n\sum_{i=1}^n \left(\frac{i}{n}\right)^k \le \frac{1}{(k+1)n^{k+1}}\left[\prod_{\ell=0}^k(n+\ell) - \prod_{\ell=0}^k\ell\right] = \frac{1}{k+1}\prod_{\ell=0}^{k}\left(1 + \frac{\ell}{n}\right)$$ Taking limsup on both sides, we obtain

$$\limsup_{n\to\infty}\frac1n\sum_{i=1}^n \left(\frac{i}{n}\right)^k \le \frac1{k+1}\limsup_{n\to\infty}\prod_{\ell=0}^k\left(1 + \frac{\ell}{n}\right) = \frac{1}{k+1}$$ By a similar argument, when $n \ge k$, we have

$$\frac1n\sum_{i=1}^n\left(\frac{i}{n}\right)^k \ge \frac1n\sum_{i=k}^n\left(\frac{i}{n}\right)^k = \frac1n\sum_{i=1}^{n-k+1}\left(\frac{i+k-1}{n}\right)^k \ge \frac{1}{n^{k+1}}\sum_{i=1}^{n-k+1}\prod_{\ell=0}^{k-1}(i+\ell)\\ = \frac{1}{(k+1)n^{k+1}} \prod_{\ell=0}^k (n-k+1+\ell) = \frac{1}{k+1}\prod_{\ell=0}^k\left(1 - \frac{k-1-\ell}{n}\right) $$ Taking liminf on both sides, we get

$$\liminf_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac{i}{n}\right)^k \ge \frac{1}{k+1}\liminf_{n\to\infty}\prod_{\ell=0}^k\left(1 - \frac{k-1-\ell}{n}\right) = \frac{1}{k+1}$$ Since both limsup and liminf of $\frac1n\sum\limits_{i=1}^n\left(\frac{i}{n}\right)^k$ exists and equals to $\frac{1}{k+1}$, by squeezing, we find:

$$\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac{i}{n}\right)^k = \frac{1}{k+1}$$

achille hui
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Using generalize harmonic numbers $$S_n=\sum_{i=1}^{n}\Bigl(\frac{i}{n}\Bigr)^k=\frac1 {n^k}\sum_{i=1}^{n}i ^k=\frac1 {n^k} H_n^{(-k)}$$ Using asymptotics $$H_n^{(-k)}=n^k \left(\frac{n}{k+1}+\frac{1}{2}+\frac{k}{12 n}+O\left(\frac{1}{n^3}\right)\right)+\zeta (-k)$$ So $$\frac 1 n S_n=n^{-k-1} \zeta (-k)+\left(\frac{1}{k+1}+\frac{1}{2 n}+\frac{k}{12 n^2}+O\left(\frac{1}{n^4}\right)\right)\to \frac{1}{k+1}$$

Claude Leibovici
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If you want to prove $$\int_0^x u^kdu={x^{k+1}\over k+1}$$I think you would better prove that $$\int_0^x f(u)du=F(x)$$by proving that $$f(x)=\lim_{h\to 0}{\int_{x}^{x+h}f(u)du\over h}$$

Mostafa Ayaz
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As per the comment I posted a while ago (considering the question was updated), using Riemann sum $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum\limits_{i=1}^n f\left(\frac{i}{n}\right)= \int\limits_{0}^{1} f(x)dx$$ we have $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\Bigl(\frac{i}{n}\Bigr)^k =\int\limits_{0}^{1}x^kdx=\frac{x^{k+1}}{k+1}\Bigg\rvert_{0}^{1} =\frac{1}{k+1}$$

rtybase
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