The question is, for any sequence of random variables $X_n$ with $\mathbb{E}[X_n^p]\rightarrow0$ for each $p\in \mathbb{N}$, can we prove (or find a counterexample to the claim that) $$ \frac{ \mathbb{E}[ X_n ] \ \mathbb{E}[ X_n^3 ]^2 }{ Var( X_n )^3 } \rightarrow 0 ? $$
I've tried a lot of combinations of the standard inequalities like Holder's, Jensen's, Markov's, and Chebyshev's to try and get an upper bound, but no luck so far for a general proof.
Also, working with the common probability distributions (e.g. Poisson, Normal, Exponential) there doesn't seem to be an obvious counter example.
I feel like the key might to be to show that $\frac{ \mathbb{E}[ X_n^3 ]^2 }{ Var( X_n )^3 }$ is bounded by some constant?
This is an extra-curricular question set by my lecturer - I'm happy to see full solutions or general hints towards the answer.
Consider the following distribution for $X_n$: $$ X_n = \begin{cases} \frac{1}{\sqrt[3]{n}}&\text{with probability} \frac{1}{n} \\ \frac{2}{\sqrt[3]{n}}&\text{with probability} \frac{n-2}{n} \\ \frac{3}{\sqrt[3]{n}}&\text{with probability} \frac{1}{n} \end{cases} $$ We have $$ \mathsf{E}(X_n^p) \leq \left(\frac{3}{\sqrt[3]{n}}\right)^p $$ and therefore, when $n \rightarrow \infty$, $\mathsf{E}(X_n^p) \rightarrow 0$.
In addition, we have $$ \mathsf{E}(X_n) = \frac{2}{\sqrt[3]{n}}, \quad \mathsf{E}(X_n^3) = \frac{8n + 12}{n^2} \quad\text{and}\quad \mathsf{Var}(X_n) = \frac{2}{n^{5/3}} $$ Therefore, $$ \frac{\mathsf{E}(X_n)\mathsf{E}(X_n^3)^2}{\mathsf{Var}(X_n)^3}=\frac{\frac{2}{\sqrt[3]{n}}\cdot \left(\frac{8n+12}{n^2}\right)^2}{\frac{8}{n^5}} \geq 16n^{8/3} $$ and it is a counterexample.