Expectation of an infinite sum of indicator functions

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Let $(X_n)$ be i.i.d. with $E|X_n| = \infty$, $E$ being expected value. I'm trying to understand a proof that ${\lim \sup}_{n \to \infty} |X_n|/n = \infty$.

In the proof the following inequality is used: for all $a \in \mathbb R$,

$$E\left[\sum_{n=1}^\infty \mathbb 1_{\frac{|X_1|}{n} > a}\right] \geq E\left[\frac{|X_1|}{a} - 1\right]$$

How should I understand this inequality? Where does it come from?

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We observe that $$\begin{eqnarray} \sum_{n=1}^\infty 1_{\{|X_1|/n>a\}}&=&\sum_{n=1}^\infty 1_{\{n<|X_1|/a\}}\\ &=&\left\lceil\frac{|X_1|}{a}\right\rceil-1\\&\ge& \frac{|X_1|}{a}-1, \end{eqnarray}$$ where $\lceil x\rceil$ denotes the least integer $\geqslant x$. Therefore it follows $$ \Bbb E\left[\sum_{n=1}^\infty 1_{\{|X_1|/n>a\}}\right]\ge \Bbb E\left[\frac{|X_1|}{a}-1\right]. $$