Showing $xP(|X|>x) \rightarrow 0$ as $x\rightarrow \infty$

Question

$X$ is a discrete random variable with distribution function $$P(X=(-1)^k k)=\frac{c}{k^2\log k}$$ for integer $k\geq 2$ and some normalization constant $c$. Is that true that $xP(|X|>x) \rightarrow 0$ as $x\rightarrow \infty$? I am having difficulties to find an appropriate upper bound that goes to zero.

Thanks!

Answer 1

We have $$ P(|X|>x) = \sum_{k>x} \frac{c}{k^2\log k} \sim \int_{x}^\infty \frac{c}{k^2\log k}dk < \frac{c}{\log x}\int_x^\infty \frac{1}{k^2}dk =\frac{c}{x\log x}$$