# Durrett Probability 5th edition: Theorem 2.3.7

The theorem 2.3.7 states the following.

If $$X_1, X_2, \ldots$$ are i.i.d with $$E \left| X_i \right| = \infty$$, then $$P(\left|X_n\right| \geq n \;\textit{i.o.}) = 1$$. So if $$S_n = X_1 + \ldots + X_n$$, then $$P(\lim \frac{S_n}{n} \textit{exists} \in (-\infty, \infty)) = 0$$.

where i.o. stands for infinitely often, or limsup of sequence of sets.

I do not understand the second part of the proof.

It lets $$C = \{\omega \colon \lim \frac{S_n}{n} \text{exists} \in (-\infty, \infty) \}$$ and states that on $$C \cap \{\omega \colon \left|X_n\right| \geq n \}$$, $$\left| \frac{S_n}{n} - \frac{S_{n+1}}{n+1} \right| > \frac{2}{3}$$ i.o.

I tried using the inequality $$\left|a-b\right| \geq \left| \left|a\right| - \left|b\right| \right|$$, but at best I could only replace $$S_{n+1}/(n+1)$$ with $$1$$, which I don't think helps much.

So if anyone could explain why $$\left| \frac{S_n}{n} - \frac{S_{n+1}}{n+1} \right| > \frac{2}{3}$$ i.o., I would appreciate that, thank you.

By algebra, \begin{align*} \big|\frac{S_{n+1}}{n+1}-\frac{S_n}{n}\big|&=\big|\frac{X_{n+1}}{n+1}+S_n(\frac{1}{n+1}-\frac{1}{n})\big|\\ &=\big|\frac{X_{n+1}}{n+1}-\frac{S_n}{n(n+1)}\big|\\ &\ge \big|\frac{X_{n+1}}{n+1}\big|-\big|\frac{S_n}{n(n+1)}\big|. \end{align*} On $$\{X_n\ge n\ \text{i.o.}\}$$, the first term $$\ge 1$$ infinitely often, and the second term goes to zero, so in particular, for all $$n$$ large enough, $$|\frac{S_n}{n(n+1)}|<1/3$$. Hence $$\big|\frac{X_{n+1}}{n+1}\big|-\big|\frac{S_n}{n(n+1)}\big| > 2/3\quad\text{infinitely often}.$$
• the choice of $2/3$ is arbitrary?
• @Phil: Yes, the choice of 2/3 is completely arbitrary, because the point of what is being shown is to show that $S_n/n$ is not a Cauchy sequence. Any fixed $\epsilon_0>0$ would work instead of $1/3$. Aug 11, 2022 at 22:47