How to evaluate $\lim_{n \to \infty} \sum_{j=1}^n |f(\frac{j}{n}) - f(\frac{j-1}{n})|^{3/2}$?

Question

Suppose $f(x)=x^3$.

How to evalueate $\lim_{n \to \infty} \sum_{j=1}^n |f(\frac{j}{n}) - f(\frac{j-1}{n})|^{3/2}$?

$$\lim_{n \to \infty} \sum_{i=1}^n |f(\frac{i}{n}) - f(\frac{i-1}{n})|^{3/2} = \lim_{n \to \infty} \sum_{j=1}^n |(\frac{j}{n})^3 - (\frac{j-1}{n})^3|^{3/2}$$

I tried $x = j/n$, so $$\left|(\frac{j}{n})^3 - (\frac{j-1}{n})^3\right|^{3/2} = (x^3-(x-\frac{1}{n})^3)^{3/2}$$

Then, I think I might need to use the integral: $$\int_0^1 (x^3-(x-\frac{1}{n})^3)^{3/2} dx$$

Am I right? I do not know how to calculate this integral.

Answer 1

The answer is $0$. Here is general result. Suppose $f$ is a function which satisfies the inequality $|f(x)-f(y)|\le C|x-y|$ for some finite constant $C$. [ Any function with a bounded derivative satisfies this; $x^{3}$ certainly does this on $[0,1]$]

Then $\sum_{j=1}^n |f(\frac{j}{n}) - f(\frac{j-1}{n})|^{3/2} \le C^{3/2}n(\frac 1 {n^{3/2}})\to 0$.