How to show that $P(X_{n} \geq n$ $i.o.) = 0$, given $E(X_{i}) = 0$ and $E((X_{i})^{2})=1$ for $i=1,2,3...$

Question

I'm struggling with the following problem (Exercise 4.5.16 in Rosenthal's probability book):

Let $X_{1}, X_{2},...$ be defined jointly on some probability space, with $E(X_{i}) = 0$ and $E((X_{i})^{2})=1$ for all $i$. Prove that $P(X_{n} \geq n$ $ i.o) = 0$.

My idea was to use the Borel-Cantelli Lemma and argue that $\sum P(X_{n} \geq n) < \infty$, but I'm stuck trying to estimate $P(X_{n} \geq n)$.

Answer 1

Since $X_n$ is centered, by Chebychev inequality we have $$ \mathrm{Pr}[X_n\ge n] \le \frac{\mathrm{Var}(X_n)}{n^2}=\frac{1}{n^2}. $$ Then we obtain $$ \sum_{n\ge 1}\mathrm{Pr}[X_n\ge n]\le \sum_{n\ge 1}\frac{1}{n^2}=\zeta(2)<\infty. $$ The claim follows by Borel Cantelli Lemma I.