Sequence of distinct moments of $X_{n}$ converging to $1$ implies $X_{n}$ converges to $1$ Suppose for $0<\alpha<\beta$ and $X_{n}\geq0$ we have $EX_{n}^{\alpha},EX_{n}^{\beta}\to1$ as $n\to\infty$.  Show that $X_{n}\to1$ in probability.  In special cases this is pretty clear (for instance, assuming $\alpha\geq1$ and $X_{n}\geq1$).  But I don't know how to prove this in the general case.  It is interesting because the question appears in a section on weak convergence (convergence in distribution), yet the problem appears to have nothing to do with this.

Define $Y_n:=X_n^\alpha$ and $p:=\beta/\alpha\gt 1$. The assumptions give that the sequence $\{Y_n,n\geqslant 1\}$ is tight, so it's enough to prove that each subsequence converges in distribution to the constant $1$. 
Take $\{Y_{n'}\}$ a subsequence which converges in distribution to $Y$· 
Since $\{Y_n,n\geqslant 1\}$ is bounded in $\mathbb L^{(1+p)/2}$, we have 
$$\mathbb EY^{(1+p)/2}=\lim_{n'\to\infty}\mathbb E(Y_{n'}^{(1+p)/2})=1.$$
We also that $\mathbb E(Y)=1$, hence by the equality case in Hölder's inequality, $Y=1$ almost surely.
