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Let $(X_n)_{n \geq 1}$ be a sequence of non-negative i.i.d random variables.
Show that $$E[X_1]\leq \sum_{n \in N} P(X_1 > n) $$
Further, show that if $E[X_1]=\infty$ then $P(X_n > n, i.o) = 1$ and $\frac{S_n} {n}$ converges almost always to $\infty$
I think the second part follows from the strong law of large numbers but I don't know how to proof the 1st one. I tried using Markov's inequality but it didn't seem to work.
The first part was already answered by @RobertW.
The second part directly follows from the second Borel-Cantelli lemma since $ \sum_{n \in N} P(X_n > n) = \sum_{n \in N} P(X_1 > n) \ge E[X_1]=\infty.$
For the third part regarding $\frac{S_n}{n}$, you can use Kronecker's lemma to create a contradiction. Suppose $\frac{S_n} {n}$ is finite for some value $\omega \in \{x_n > n\ \ i.o.\}$. Then we can write $\lim_{n\rightarrow \infty}\sum_{m=1}^n\frac{x_m(\omega)}{n}=s$ where $s$ is a real value. By Kronecker's Lemma, it follows that $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{m=1}^nm\frac{x_m(\omega)}{n}=0.$ But this is a contradiction since $ \frac{1}{n}\sum_{m=1}^nm\frac{x_m(\omega)}{n}\ge \frac{1}{n}n\frac{x_n(\omega)}{n}=\frac{x_n(\omega)}{n} > 1 $ i.o.
