Tight sequence of rv's such that $V(X_n) \rightarrow +\infty$. 
Let $(X_n)$ be a tight sequence of real valued rv's, i.e. $\displaystyle \lim_K\sup_n P\left(\left|X_n\right|>K\right)=0$, defined on a common probability space, such that $E\left(X_n^2\right)<+\infty$  and $E\left(X_n^2\right) \rightarrow +\infty.$ Prove that $V\left(X_n\right) \rightarrow +\infty$.

Attempt Since $V(X_n)=E\left(X_n^2\right)-E\left(X_n\right)^2$, we just have to show that $(E(X_n))$ does not converge to $+\infty$. Obviously, we need the tightness property for the last point but i don't seem how to get the result.
Thanks a lot!
 A: In this setting, it may be not true that $\mathbb E\left[X_n\right]$ does not go to infinity. For example, if $\left(A_n\right)_{n\geqslant 1}$ is a sequence of measurable set such that $\mathbb P\left(A_n\right)=1/n^2$ and $X_n=n^3\mathbf 1\left(A_n\right)$, then we have tightness and $\mathbb E\left[X_n^2\right]=n^4$ but $\mathbb E\left[X_n\right]=n$ which goes to infinity. 
In order to solve the problem, we can show that $$\tag{*}\lim_{n\to +\infty}\frac{\left(\mathbb E\left[X_n\right]\right)^2}{\mathbb E\left[X_n^2\right]}=0.$$
This can be done in the following way:
\begin{align}\left|\mathbb E\left[X_n\right]\right|&\leqslant 
R+\mathbb E\left[\left|X_n\right|\mathbf 1\left\{\left|X_n\right|\gt R\right\}\right]\\
&\leqslant R+\left(\mathbb E\left[X_n^2\right]\right)^{1/2}\left(\mathbb P\left\{\left|X_n\right|\gt R\right\}\right)^{1/2}.
\end{align}
Then use the elementary inequality $(a+b)^2\leqslant 2a^2+2b^2$, divide by $\mathbb E\left[X_n^2\right]$ and take the $\limsup_{n\to +\infty}$. Then conclude by tightness.
