For $r>s\geq1$, convergence in $s^{\text{th}}$ mean does not imply convergence in $r^{\text{th}}$ mean

Question

I need a counterexample for the problem: if $r>s\geq1$, convergence in $s^{\text{th}}$ mean does not imply convergence in $r^{\text{th}}$ mean.

The definition for convergence in mean is as follows: Let $r\geq1$ be a fixed number. A sequence of random variables $X_1, X_2, X_3,...$ converges in the $r^{\text{th}}$ mean to a random variable $X$, if

$$\lim_{n\to\infty}E(|X_n−X|^r)=0.$$

Answer 1

Try $P(X_n=n)=1/n^\alpha$ and $P(X_n=0)=1-P(X_n=n)$ with $X=0$.

Then, $$ E|X_n-X|^s=EX_n^s=n^{s-\alpha} $$