Let $(X_n)_n$ be a sequence of identically distributed random variables with $\mathbb{E}X_1^2 < \infty$. Show that $$\lim_{n \to \infty} n\mathbb{P}\{|X_1| \geq \epsilon \sqrt{n}\} = 0$$ for all $\epsilon > 0$.
Here is the solution my teaching assistant provides:
We know that
$$\int_0^\infty \mathbb{P}\{X_1^2 \geq t\epsilon^2\}dt = \mathbb{E}\left[\frac{X_1^2}{\epsilon^2}\right]< \infty$$
Hence $$\sum_{n=1}^\infty \mathbb{P}\{X_1^2 \geq n \epsilon^2\} = \sum_{n=1}^\infty \mathbb{P}\{|X_1| \geq \sqrt{n} \epsilon\}< \infty$$
and because $\sum_{n=1}^\infty \frac{1}{n}= \infty$, it follows that $n \mathbb{P}\{|X_1| \geq \epsilon \sqrt{n}\} \to 0$ when $n \to \infty$.
Questions:
How do we deduce the last step? (I.e. $n \mathbb{P}\{|X_1| \geq \sqrt{n}\epsilon\} \to 0$ from divergence of the harmonic series?)
I do not think this is correct. I think we can fix the argument with the Borel Cantelli lemma though.
Indeed, it seems that what is written is that "if $a_n$, is a sequence of positive numbers such that $\sum_n a_n/n$ is convergent, then $a_n\to 0$. But this may not be true, for example if $a_n=1$ when $n$ has the for $2^k$ for some $k\geqslant 1$ and zero otherwise. However, it is true if there exists positive constants $c$ and $C$ such that for all $N\geqslant 1$, $$ ca_{2^{N+1}}\leqslant \min_{2^N\leqslant n\leqslant 2^{N+1}}a_n\mbox{ and } \max_{2^N\leqslant n\leqslant 2^{N+1}}\leqslant Ca_{2^{N}} . $$