Proof that the series $\sum\limits_n P\left(\max\limits_{1\leq k\leq n}|X_{nk}|>\theta\sqrt{n\log n}\right)$ converges 
Let $\{X_{nk}\}$ be an array of i.i.d. random variables such that$$E|X_{11}|^4(\log^+|X_{11}|)^{-2}<\infty,\qquad EX_{11}^2=1,\qquad EX_{11}=0.$$ Then the series $$S=\sum_{n=1}^\infty P \left(\max_{1\leq k\leq n}|X_{nk}|>\theta\sqrt{n\log n} \right)$$ converges.

This is what I saw in a paper and I failed to prove it. By Chebyshev inequality, 
\begin{equation}\nonumber
\begin{aligned}
S \leq&\sum_{n=1}^\infty E\left[\max_{1\leq k\leq n}|X_{nk}|^4(\log^+|X_{nk}|)^{-2}\right]
\frac{\left(\log\theta\sqrt{n\log n}\right)^2}{\theta^4n^2\log^2n}\\
\le&\sum_{n=1}^\infty E\left[|X_{11}|^4(\log^+|X_{11}|)^{-2}\right]
\frac{\left(\log\theta+\frac{1}{2}\log n+\frac{1}{2}\log\log n\right)^2}{\theta^4n\log^2n}.\\
\end{aligned}
\end{equation}
However, the above series is divergent. 

How to prove that $S$ converges?

 A: Since the random variables $X_{n,k}$ have the same distribution, it suffices to prove that for each positive $\theta$, the series $\sum_n  n\Pr\left(Y\gt \theta\sqrt{n\log n}\right)$ converges, where $Y$ is a random variable having the same distribution as $\left|X_{1, 1}\right|$. Since for $n$ such that $2^N\leqslant n\lt 2^{N+1}$, inequality $$n\Pr\left(Y\gt \theta\sqrt{n\log n}\right)\leqslant 2^{N+1} \Pr\left(Y\gt \theta 2^{N/2}\sqrt{N\log 2}    \right), $$
 it suffices to prove the convergence of  the series
$$\sum_{N=1}^{+\infty}  2^{2N} \Pr\left(Y^2\gt \theta 2^{N}N   \right)  $$
for each positive $\theta$. We define $A_l:=\left\{\theta 2^l l\lt Y^2\leqslant\theta 2^{l+1}(l+1)\right\}$. Then 
$$\sum_{N=1}^{+\infty}  2^{2N} \Pr\left(Y^2\gt \theta 2^{N}N   \right)
=\sum_{N=1}^{+\infty}  2^{2N} \sum_{l \geqslant N}  \Pr\left(A_l \right)= \sum_{ l =1} ^{ +\infty}\Pr\left(A_l \right)\sum_{N=1}^l2^{2N}    \leqslant  \sum_{ l =1} ^{ +\infty}4^{l+1} \Pr\left(A_l \right) .$$
Now, use the fact that 
$$\theta^2 \frac{ 2^{2l}l^2}{\left(\log^+\left(\theta 2^l l\right)\right)^2}  \mathbf 1\left(A_l\right) \leqslant \frac{Y^4 }{\left(\log^+ Y\right)^2}  \mathbf 1\left(A_l\right)     $$
to get the wanted result.
