Darboux sums: $f(x)=x^k $ on $[0,1]$

Question

Given $k>0$, define $f \colon [0,1] \to \mathbb{R}$ by $f(x)=x^k$. Let $P_n=\{0,\frac{1}{n},\frac{2}{n},...,1\}$ be the regular $n$- partition of the interval $[0,1]$. Prove that $L(f,P_n) \le \frac{1}{k+1} \le U(f,P_n).$ Hint: Don't think or work too hard.

Could someone give me a small push in the right direction? I was thinking of starting with the upper and lower darboux sums but I get a summation of the following form $\sum_{i=1}^{n}\frac{1}{n^{k+1}} i^k$ so I doubt the problem wants us to head that way and there must be a nicer trick as the hint suggests.

Answer 1

For $k \in \mathbb{N}$ you can use

$$\frac{i^{k+1} - (i-1)^{k+1}}{k+1} = [i - (i-1)]\frac{i^k + i^{k-1}(i-1) + \ldots + (i-1)^k}{k+1},$$

Since $(k+1)(i-1)^{k} < i^k + i^{k-1}(i-1) + \ldots + (i-1)^k < (k+1)i^k$, this implies

$$(i-1)^k < \frac{i^{k+1} - (i-1)^{k+1}}{k+1}< i^k.$$

Dividing by $n^{k+1}$ and summing from $i=1$ to $i=n$ we get

$$\frac{1}{n^{k+1}}\sum_{i=1}^n(i-1)^k < \frac{1}{n^{k+1}} \underbrace {\sum_{i=1}^n\frac{i^{k+1} - (i-1)^{k+1}}{k+1}}_{\text {telescoping sum}= \frac{n^{k+1}}{k+1}} = \frac{1}{k+1}< \frac{1}{n^{k+1}}\sum_{i=1}^ni^k.$$

Answer 2

You can prove a version of the desired inequalities at each of the points in the partition separately. Specifically, you know

$$\int_{j/n}^{(j+1)/n} x^k dx = \frac{((j+1)/n)^{k+1}-(j/n)^{k+1}}{k+1}$$

and so you also know that this quantity is between $j^k/n^{k+1}$ and $(j+1)^k/n^{k+1}$, since these are the lower and upper sums of this integral with a trivial partition containing just one interval. If you can prove these bounds by some other method (without relying on knowing the answer), then you can sum them over $j$ to get the desired result. The summation over $j$ is quite simple in the end because the sum in the middle is telescoping.

One way to prove the desired inequalities at each point in the partition separately is to apply the mean value theorem to this "intermediate" quantity, which will tell you that this whole thing is given by $\frac{1}{n} z^k$ for some $z$ between $j/n$ and $(j+1)/n$.