Calculating bounds for a certain expression using Faulhaber Let $F(n,k)=\sum_{i=1}^ni^k$ for $k\geq 0$ and $n\geq 1$. I would like to prove
$$\sum_{i=1}^{n-1}i^k\leq\frac{n^{k+1}}{k+1}\leq\sum_{i=1}^ni^k.$$
The idea is to use integration where the upper and lower bounds becomes upper and lower sums of some function $g(x)$.
The middle term suggests using the integral $\int_0^n x^kdx$, i.e. letting $g(x)=x^k$.
How to choose partition $P=(a_0,\ldots,a_n)$ of $[0,n]$ so that $U(g,P)=\sum_{i=1}^ni^k$ and $L(g,P)=\sum_{i=1}^{n-1}i^k$?
 A: When $k \geqslant 0$, $x \mapsto x^k$ is increasing for $x >0$ and  if $\frac{i-1}{n} \leqslant x \leqslant \frac{i}{n}$ then $\left(\frac{i-1}{n}\right)^k\leqslant x^k \leqslant \left(\frac{i}{n}\right)^k$.
Integrating over $\left[\frac{i-1}{n},\frac{i}{n}\right]$,  yields
$$\tag{1}\frac{(i-1)^k}{n^{k+1}}= \left(\frac{i-1}{n}\right)^k\cdot \frac{1}{n} \leqslant \int_{\frac{i-1}{n}}^{\frac{i}{n}}x^k \, dx \leqslant \left(\frac{i}{n}\right)^k\cdot \frac{1}{n} =\frac{i^k}{n^{k+1}},$$
In summing (1)  from $i=1$ to $i=n$, the lower and upper sums appear and we get
$$\tag{2}\frac{1}{n^{k+1}}\sum_{i=1}^n (i-1)^k = \frac{1}{n}\sum_{i=1}^n\left(\frac{i-1}{n}\right)^k \leqslant \int_0^1 x^k \, dx \leqslant \frac{1}{n}\sum_{i=1}^n\left(\frac{i}{n}\right)^k = \frac{1}{n^{k+1}}\sum_{i=1}^n i^k$$
The desired result is obtained by substituting into (2) with
$$\sum_{i=1}^n (i-1)^k = \sum_{i=1}^{n-1}i^k, \quad \int_0^1 x^k \, dx = \frac{1}{k+1}$$
Whence,
$$\frac{1}{n^{k+1}}\sum_{i=1}^{n-1}i^k \leqslant \frac{1}{k+1} \leqslant \frac{1}{n^{k+1}}\sum_{i=1}^n i^k$$
