If $\{X_n\}$ is a sequence of iid random variables such that $\mathbb P(X_1=\pm j)=\frac c{j^2\log j}$, $j=3,4,\dots$, where $c$ is the appropriate normalizing constant.
Show that for each $M>0$, we have $\mathbb P(|X_n|>Mn \;\text{ i.o.})=1$
I am trying to use Borel-Cantelli. To do so, we need to prove that the following double sum is divergent- $$\sum_{n=1}^\infty \sum_{k=\min\{3,\lfloor Mn\rfloor+1\}}^\infty \frac 1{k^2\log k}$$ I know that $$\lim_{n\to \infty}n\sum_{k=n}^\infty \frac 1{k^2\log k}=0$$ and I can prove that $$\sum_{k=1}^\infty \frac 1{k^2\log k}<\infty$$ using Comparison Test with $\sum_{n=1}^\infty\frac 1{n^2}$.
But, I can't handle the problem at hand. Is there some other way to solve the original problem?
A hint given in the question is to find an appropriate sequence $\{a_N\}$ such that $\mathbb P(|X_1|>N)\sim a_N$ as $N\to \infty$, although I don't know how to use this hint.
What you need is to lower bound $$ \sum_{k=l}^\infty \frac{1}{k^2\log k} $$
We can lower bound by the normal rectangular argument, i.e. $\frac{1}{k^2\log k} \geq \frac{1}{x^2\log x}$ on $x\in [k, k+1)$. Note, here that we don't need to integrate on whole $[l, \infty)$ to obtain a lower bound, i.e.
$$ \sum_{k=l}^\infty \frac{1}{k^2\log k} \geq \int_l^{2l}\frac{dx}{x^2\log x} \geq \frac{1}{\log (2l)}\int_l^{2l}\frac{dx}{x^2} = \frac{1}{2l \log(2l)} $$
Using, this we can roughly say (I believe you can make it accurate)
$$ \sum_{n=1}^\infty P(|X_n|>Mn) > \sum_{n=1}^\infty \frac{1}{2Mn \log (2Mn)} $$
We can check the RHS$=\infty$ by relating it to $\int_c^\infty \frac{dx}{x\log x}=\infty$.