Let $X$ be a random variable. For any $r>0$, $\mathscr{E}|X|^r<\infty$, we have $\mathscr{E}|X|^r<\infty$ if and only if $\sum_{n=1}^\infty n^{r-1}\mathscr{P}\{|X|\geq n\}$ converges.
$\bullet$ Proof.
Given $r>0$, define series of sets $\Lambda_n=\{n\leq|X|<n+1\}$, and since $\Lambda_n$'s are disjoint, we have, by additivity property $$\mathscr{E}(|X|^r)=\int_{\cup_n\Lambda_n}|X|^rd\mathscr{P}=\sum_n\int_{\Lambda_n}|X|^rd\mathscr{P}.$$Thus, by mean value theorem, for each $\Lambda_n$, $$\begin{array}{rl}
\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)\leq\mathscr{E}(|X|^r) & \leq \sum_{n=0}^\infty (n+1)^r\mathscr{P}(\Lambda_n) \\
& = \sum_{n=0}^\infty\left[\sum_{k=0}^{r}{r\choose k}n^k\mathscr{P}(\Lambda_n)\right] \\
& = \sum_{k=0}^{r}\left[{r\choose k}\sum_{n=0}^\infty n^k\mathscr{P}(\Lambda_n)\right] \\
& = 1 + \sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n) + \sum_{k=1}^{r-1}\left[{r\choose k}\sum_{n=0}^\infty n^k\mathscr{P}(\Lambda_n)\right].
\end{array}$$Since $\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)$ dominates all the terms of $\sum_{n=0}^\infty n^k\mathscr{P}(\Lambda_n)$, $k<r$, on the right hand side of the inequality, we only need to show that the convergence equivalency between $$\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)\mbox{ and }\sum_{n=0}^\infty n^{r-1}\mathscr{P}(|X|\geq n).$$For $N\geq1$ and finite, $$\begin{array}{rl}
\sum_{n=0}^N n^r\mathscr{P}(\Lambda_n) & = \sum_{n=0}^N n^r\{\mathscr{P}(|X|\geq n)-\mathscr{P}(|X|\geq n+1)\} \\
& = \sum_{n=1}^N \left[n^r-(n-1)^r\right]\mathscr{P}(|X|\geq n) - N^r\mathscr{P}(|X|\geq N+1) \\
& = r\sum_{n=1}^N n^{r-1}\mathscr{P}(|X|\geq n) +
\sum_{k=0}^{r-2}\left[{r\choose k}(-1)^{r-k+1}\sum_{n=1}^N n^k\mathscr{P}(|X|\geq n)\right] \\
& \quad - N^r\mathscr{P}(|X|\geq N+1).
\end{array}$$We have $$\begin{array}{rl}
\sum_{n=0}^N n^r\mathscr{P}(\Lambda_n) & \leq
r\sum_{n=1}^N n^{r-1}\mathscr{P}(|X|\geq n) + \sum_{k=0}^{r-2}\left[{r\choose k}(-1)^{r-k+1}\sum_{n=1}^N n^k\mathscr{P}(|X|\geq n)\right] \\
& \leq \sum_{n=0}^N n^r\mathscr{P}(\Lambda_n)+N^r\mathscr{P}(|X|\geq N+1),
\end{array}$$and by mean value theorem $$N^r\mathscr{P}(|X|\geq N+1)=N^r\int_{|X|\geq N+1}d\mathscr{P}\leq\int_{|X|\geq N+1}|X|^rd\mathscr{P}.$$If $\mathscr{E}(|X|^r)<\infty$, then $ N^r\mathscr{P}(|X|\geq N+1) \rightarrow 0$ as $N\rightarrow\infty$. We have $\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)\Rightarrow
\sum_{n=0}^\infty n^{r-1}\mathscr{P}(|X|\geq n)<\infty$ and naturally $\sum_{n=1}^\infty n^k\mathscr{P}(|X|\geq n)<\infty$, for $k<r-1$.
If $\mathscr{E}(|X|)=\infty$, then $\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n) = \infty$ leads to $\sum_{n=1}^\infty n^k\mathscr{P}(|X|\geq n)=\infty$ for some $k\leq r-1$, hence $\sum_{n=0}^\infty n^{r-1}\mathscr{P}(|X|\geq n)=\infty.$
\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)\leq\mathscr{E}(|X|^r) & \leq \sum_{n=0}^\infty (n+1)^r\mathscr{P}(\Lambda_n) \\
& = \sum_{n=0}^\infty\left[\sum_{k=0}^{r}{r\choose k}n^k\mathscr{P}(\Lambda_n)\right] \\
& = \sum_{k=0}^{r}\left[{r\choose k}\sum_{n=0}^\infty n^k\mathscr{P}(\Lambda_n)\right] \\
& = 1 + \sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n) + \sum_{k=1}^{r-1}\left[{r\choose k}\sum_{n=0}^\infty n^k\mathscr{P}(\Lambda_n)\right].
\end{array}$$Since $\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)$ dominates all the terms of $\sum_{n=0}^\infty n^k\mathscr{P}(\Lambda_n)$, $k<r$, on the right hand side of the inequality, we only need to show that the convergence equivalency between $$\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)\mbox{ and }\sum_{n=0}^\infty n^{r-1}\mathscr{P}(|X|\geq n).$$For $N\geq1$ and finite, $$\begin{array}{rl}
\sum_{n=0}^N n^r\mathscr{P}(\Lambda_n) & = \sum_{n=0}^N n^r\{\mathscr{P}(|X|\geq n)-\mathscr{P}(|X|\geq n+1)\} \\
& = \sum_{n=1}^N \left[n^r-(n-1)^r\right]\mathscr{P}(|X|\geq n) - N^r\mathscr{P}(|X|\geq N+1) \\
& = r\sum_{n=1}^N n^{r-1}\mathscr{P}(|X|\geq n) +
\sum_{k=0}^{r-2}\left[{r\choose k}(-1)^{r-k+1}\sum_{n=1}^N n^k\mathscr{P}(|X|\geq n)\right] \\
& \quad - N^r\mathscr{P}(|X|\geq N+1).
\end{array}$$We have $$\begin{array}{rl}
\sum_{n=0}^N n^r\mathscr{P}(\Lambda_n) & \leq
r\sum_{n=1}^N n^{r-1}\mathscr{P}(|X|\geq n) + \sum_{k=0}^{r-2}\left[{r\choose k}(-1)^{r-k+1}\sum_{n=1}^N n^k\mathscr{P}(|X|\geq n)\right] \\
& \leq \sum_{n=0}^N n^r\mathscr{P}(\Lambda_n)+N^r\mathscr{P}(|X|\geq N+1),
\end{array}$$and by mean value theorem $$N^r\mathscr{P}(|X|\geq N+1)=N^r\int_{|X|\geq N+1}d\mathscr{P}\leq\int_{|X|\geq N+1}|X|^rd\mathscr{P}.$$If $\mathscr{E}(|X|^r)<\infty$, then $ N^r\mathscr{P}(|X|\geq N+1) \rightarrow 0$ as $N\rightarrow\infty$. We have $\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n)\Rightarrow
\sum_{n=0}^\infty n^{r-1}\mathscr{P}(|X|\geq n)<\infty$ and naturally $\sum_{n=1}^\infty n^k\mathscr{P}(|X|\geq n)<\infty$, for $k<r-1$.
If $\mathscr{E}(|X|)=\infty$, then $\sum_{n=0}^\infty n^r\mathscr{P}(\Lambda_n) = \infty$ leads to $\sum_{n=1}^\infty n^k\mathscr{P}(|X|\geq n)=\infty$ for some $k\leq r-1$, hence $\sum_{n=0}^\infty n^{r-1}\mathscr{P}(|X|\geq n)=\infty.$
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