Let $X$ be a random variable taking values in the natural numbers with $\mathbf{P}(X=n)=\frac{1}{\zeta(3)n^3}$ ($\zeta(3)\approx 1.20$).
I want to show that $X$ is absolutely integrable. In probability, I think this is the same as showing that $\mathbf{E}(X)<\infty$. $$\mathbf{E}(X)=\int_\mathbb{N}X(\omega)d\mathbf{P}=\sum_{i=1}^\infty \mathbf{P}(X=i)=\frac{1}{\zeta(3)}\sum_{i=1}^\infty\frac{1}{n^3}=\frac{\zeta(3)}{\zeta(3)}=1<\infty,$$ so $X$ is absolutely integrable.
Is this correct?