Suppose $X$ is a random variable, $a>0$ is a constant. Prove that $\mathbb{E}\vert X \vert^{a}<\infty$ iff $\sum\limits_{n=1}^{\infty}n^{a-1}\mathbb{P}(\vert X \vert \geqslant n)<\infty$.
This is a exercise from Mathematical Statistics. Jun Shao. Second edition. EX1.54
For convenience, we suppose that $X$ is non-negative.
I know how to prove in the case $a=1$, i.e. $\mathbb{E}X <\infty$ iff $\sum\limits_{n=1}^{\infty}\mathbb{P}(X \geqslant n)<\infty$.
Using the inequality below: $$ \sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\} \leqslant \mathbb{E}X \leqslant 1+\sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\} $$
This is because: \begin{align} \sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\}&=\sum_{n=1}^{\infty}\sum_{k\geqslant n}\mathbb{P}\{k\leqslant X < k+1\}=\sum_{k=1}^{\infty}k\mathbb{P}\{k\leqslant X <k+1\} \\ &=\sum_{k=0}^{\infty}\mathbb{E}[k I\{k\leqslant X <k+1\}]\leqslant \sum_{k=0}^{\infty}\mathbb{E}[XI\{k\leqslant X <k+1\}] \\ &=\mathbb{E}X \\ & \leqslant \sum_{k=0}^{\infty}\mathbb{E}[(k+1)I\{k\leqslant X <k+1\}] \\ &=\sum_{k=0}^{\infty}(k+1)\mathbb{P}\{k\leqslant X <k+1\} \\ &=\sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\}+\sum_{k=0}^{\infty}\mathbb{P}\{k\leqslant X <k+1\}=\sum_{n=1}^{\infty}\mathbb{P}\{X\geqslant n\}+1 \end{align}
So, $\mathbb{E}X <\infty$ iff $\sum\limits_{n=1}^{\infty}\mathbb{P}(X \geqslant n)<\infty$.
But for $a>0,a\neq 1$, I have no idea how to change the inequality above to help me to prove the proposition.
