I solved exercise 3.6.13 page 40 of Rosenthal's A First Look at Rigorous Probability Theory(which should appear at the end of chapter 4 about expectations). The fact that I didn't use one of its assumptions ($E(X_n)=0$ for all $n$) makes me doubt about my solution, or about the existence of an interesting different solution. Could you please check it out?
Let $X_1,X_2,\dots$ be defined jointly on some probability space $(\Omega,\mathcal{F},P)$, with $E(X_n)=0$ and $E(X_n^2)=1$ for all $n$. Prove that $P(X_n\ge n\,i.o.)=0$.
i.o. stands for infinitely often. For a sequence $\{A_n\}_n$ of events,
$$\{A_n\,i.o.\}=\limsup_nA_n=\bigcap_{i=1}^{+\infty}\bigcup_{j=i}^{+\infty}A_j.$$
Here's my solution:
The Borel-Cantelli Lemma usually comes handy in such situations. To solve the exercise, it's enough to prove that $\sum_{n=1}^{+\infty}P(X_n\ge n)$ is finite.
Let $n\in\mathbb{N}$. We have:
\begin{aligned} 1&=E(X_n^2)\\ &=\int_\Omega X_n^2\,\mathrm{d}P\\ &=\int_{\{X_n^2<n^2\}}X_n^2\,\mathrm{d}P+\int_{\{X_n^2\ge n^2\}}X_n^2\,\mathrm{d}P\\ &\ge\int_{\{X_n^2\ge n^2\}}X_n^2\,\mathrm{d}P\\ &\ge n^2P(X_n^2\ge n^2). \end{aligned}
Thus $P(X_n^2\ge n^2)\le\dfrac{1}{n^2}$. As $\{X_n\ge n\}\subset\{X_n^2\ge n^2\}$, we get $P(X_n\ge n)\le\dfrac{1}{n^2}$. Hence $\sum_{n=1}^{+\infty}P(X_n\ge n)$ is finite as desired. ($E(X_n)=0$ for all $n$ was not used)
You can use the Chebychev's inequality, which is $$P(X_{n}\geq n)\leq \frac{Var(X_{n})}{n^2}=\frac{1}{n^2}.$$ Because $Var(X_{n})=E(X_{n}^2)-(E(X_{n}))^2=1$, and the proof of this inequality is easy.