Limit of a Power Sum 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.
 A: 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:
$$\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac{i}{n}\right)^k = \frac{1}{k+1}$$
A: 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
$$\frac 1 n S_n=n^{-k-1} \zeta (-k)+\left(\frac{1}{k+1}+\frac{1}{2 n}+\frac{k}{12
   n^2}+O\left(\frac{1}{n^4}\right)\right)\to \frac{1}{k+1}$$
A: 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 $$f(x)=\lim_{h\to 0}{\int_{x}^{x+h}f(u)du\over h}$$
A: 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
$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\Bigl(\frac{i}{n}\Bigr)^k
=\int\limits_{0}^{1}x^kdx=\frac{x^{k+1}}{k+1}\Bigg\rvert_{0}^{1} 
=\frac{1}{k+1}$$
