Convergence of $E(|X|^k)$ and $\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i)$ I am asked to prove the following question regarding convergence of $k^\text{th}$ moment of a random variable:
$E(|X|^k) < \infty$ if and only if $\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i) < \infty$ for any integer $k > 1$. I have managed to prove the direction that given $\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i) < \infty$, we have $E(|X|^k) < \infty$. But I am stuck at the other direction. 
The proof I have come up with now is as follow:
Construct $X_i^k(\omega) = (i-1)^kI_{A_i}$ where $A_i = \{i-1 \leq |X| < i\}$. Then $|X|^k \geq \sum_{i=1}^{\infty}X_i^k$, and we have $E(|X|^k) \geq E(\sum_{i=1}^{\infty}X_i^k)$. Since $\sum_{i=1}^{\infty}X_i^k$ is a discrete random varaible taking value $(i-1)^k$ with probability $P(i-1 \leq |X| < i)$. We have
$$
E(|X|^k) \geq \sum_{i=1}^{\infty}(i-1)^kP(i-1 \leq |X| < i).
$$
I know the next step is to somehow show that the RHS is greater than $\sum_{i=1}^{\infty}i^{k-1}P(|X| \geq i)$, but I am stuck at this point and I don't know how to proceed. Can anyone give some hints? Thanks very much!
 A: In what follows, $k$ is any real number with $k\geq 1$ (and we also use the convention $0^0:=1$).  Since we are dealing with $|X|$ anyhow, we may very well assume that $X$ is a nonnegative random variable (otherwise, simply replace $X$ by $|X|$).  I use the notation $\mathbb{E}[Y]$ for the expected value of a random variable $Y$ and $\mathbb{P}$ is the probability measure.
From this answer, we have
$$\mathbb{E}\left[X^k\right]=\int_{[0,\infty)}\,\mathbb{P}\left[X^k> t\right]\,\text{d}t\,.$$
Write $u:=t^{\frac1k}$, and so we get
$$\mathbb{E}\left[X^k\right]=k\,\int_{[0,\infty)}\,u^{k-1}\,\mathbb{P}\left[X^k>u^k\right]\,\text{d}u=k\,\int_{[0,\infty)}\,u^{k-1}\,\mathbb{P}\left[X>u\right]\,\text{d}u\,,$$
since $X\geq 0$.  
Now,
$$\begin{align}\mathbb{E}\left[X^k\right]&=k\,\sum_{n=0}^\infty\,\int_{[n,n+1)}\,u^{k-1}\,\mathbb{P}[X>u]\,\text{d}u\\&\geq k\,\sum_{n=0}^\infty\,\int_{[n,n+1)}\,n^{k-1}\,\mathbb{P}[X\geq n+1]\,\text{d}x=k\,\sum_{n=0}^\infty\,n^{k-1}\,\mathbb{P}[X\geq n+1]\,.\tag{1}\end{align}$$
Similarly,
$$\begin{align}\mathbb{E}\left[X^k\right]&=k\,\sum_{n=0}^\infty\,\int_{[n,n+1)}\,u^{k-1}\,\mathbb{P}[X>u]\,\text{d}u\\&\leq k\,\sum_{n=0}^\infty\,\int_{[n,n+1)}\,(n+1)^{k-1}\,\mathbb{P}[X\geq n]\,\text{d}x=k\,\sum_{n=0}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n]\,.\tag{2}\end{align}$$
If $\mathbb{E}\left[X^k\right]<\infty$, then (1) implies that
$$\sum_{n=0}^\infty\,n^{k-1}\,\mathbb{P}[X\geq n+1]<\infty\,.$$
Thus, for any integer $m\geq 0$,
$$\sum_{n=m}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n]<\infty\,.$$
In particular, if $m$ is so chosen that $(n+1)^{k-1}\leq 2n^{k-1}$ for every integer $n\geq m$ (which always exists, depending on $k$), then
$$\sum_{n=m}^\infty\,\frac{(n+1)^{k-1}}{2}\,\mathbb{P}[X\geq n+1]\leq \sum_{n=m}^\infty\,n^{k-1}\,\mathbb{P}[X\geq n+1]<\infty\,.$$
This shows that
$$\begin{align}\sum_{n=1}^\infty\,n^{k-1}\,\mathbb{P}[X\geq n]&=\sum_{n=0}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n+1]
\\&=\sum_{n=0}^{m-1}\,(n+1)^{k-1}\,\mathbb{P}[X\geq n+1]+\sum_{n=m}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n+1]<\infty\,.
\end{align}$$
If $\mathbb{E}\left[X^k\right]=\infty$, then (2) implies that
$$\sum_{n=0}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n]=\infty\,.$$
Thus, for any integer $m\geq 0$,
$$\sum_{n=m}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n]=\infty\,.$$
In particular, if $m$ is so chosen that $m>0$ and $(n+1)^{k-1}\leq 2n^{k-1}$ for every integer $n\geq m$ (which always exists, depending on $k$), we get
$$\sum_{n=m}^\infty\,2n^{k-1}\,\mathbb{P}[X\geq n]\geq \sum_{n=m}^\infty\,(n+1)^{k-1}\,\mathbb{P}[X\geq n]=\infty\,,$$
so
$$\sum_{n=1}^\infty\,n^{k-1}\,\mathbb{P}[X\geq n]\geq \sum_{n=m}^\infty\,n^{k-1}\,\mathbb{P}[X\geq n]=\infty\,.$$
