In an article I found the following:
If $X$ is a r.v. with zero mean and finite variance, then $$ \sum_N \frac 1 {N^2} \mathbb E\left[ |X|^4 \mathbf 1_{|X|<\sqrt N} \right]<+\infty $$
and I am struggling to understand how to prove it. I tried to do the classical estimation, that is $$ \mathbb E\left[ |X|^4 \mathbf 1_{|X|<\sqrt N} \right] \le N \mathbb E\left[ |X|^2 \mathbf 1_{|X|<\sqrt N} \right] \le N $$ but it is not enough. I guess I can get $o(N)$, but that is still not enough.
I also tried to come up with some counterexample, but for example a continuous distribution with a density with tail $O(x^{-k})$ needs $k>3$ to have finite variance, that coincides with the condition to get summability.
And if $X$ has a distribution with compact support, then all the moments are bounded by a same constant, so the summability follows.
Ok, I probably got it.
$$ \sum_{N=1}^{\infty} \frac 1 {N^2} \mathbb E\left[ |X|^4 \mathbf 1_{|X|<\sqrt N} \right] = \sum_{N=1}^{\infty} \frac 1 {N^2} \sum_{n=1}^{N}\mathbb E\left[ |X|^4 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right]\\ = \sum_{n=1}^{\infty}\mathbb E\left[ |X|^4 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right]\sum_{N=n}^{\infty} \frac 1 {N^2} \\ \le \sum_{n=1}^{\infty}\frac 2n\mathbb E\left[ |X|^4 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right]\\ \le \sum_{n=1}^{\infty}2\mathbb E\left[ |X|^2 \mathbf 1_{\sqrt{n-1}\le|X|<\sqrt n} \right] = 2 \text{Var}(X) <+\infty $$