Consider a positive discrete random variable $X$ taking values in $\mathbb{N}$ and let $r\geq 1$. Is it true that for the $r$ moment of the expectation the following relation holds:
\begin{equation} \mathbb{E}[X^r] = \sum_{k=1}^\infty k^{r-1} \mathbb{P}\{X>k\}. \end{equation}
I tried using Abel's partial summation formula to show something similar but it did not work out. Any hints? Thanks!
That formula is not correct (just test it for simple cases as in my comment above).
Assuming that $X \in \{1, 2, 3, ...\}$ and $r \in \mathbb{R}$, I think the formula you want is: $$ X^r = 1 + \sum_{k=1}^{\infty} [(k+1)^r-k^r]1_{\{X>k\}} \implies \boxed{E[X^r] = 1 + \sum_{k=1}^{\infty} [(k+1)^r-k^r]P[X>k]}$$ where $1_{\{X>k\}}$ is an indicator function that is $1$ if $X>k$, and 0 else.
