I was studying this theorem:
If $X_1,X_2, \dots$ are i.i.d. with $E(X_1)=\infty$, then $P(|X_n|\geq n i.o)=1$. So if $S_n=X_1+\dots +X_n$ then $P(\lim S_n/n \space \text{exists}\in (-\infty , \infty))=0$.
Where this was written:
$$\frac{S_n}{n}-\frac{S_{n+1}}{n+1}=\frac{S_n}{n(n+1)}-\frac{X_{n+1}}{n+1}$$
take $C=\{ \omega : \lim_{n\to\infty} S_n/n\space \text{exists} \in (-\infty, \infty)\}$,$S_n/(n(n+1))\to 0$. So, on $C\cap \{ \omega : |X_n|\geq n\space \text{i.o.}\}$, we have$$\large{|}\frac{S_n}{n}-\frac{S_{n+1}}{n+1}\large{|}>2/3\space\space \text{i.o.}(*^{\infty})$$.
In $(*^{\infty})$ why $2/3$, how did that value came?
Note: $(*^{\infty})\space \text{implies}\space (***\dots)$
i.o.= infinitely often
On $C$, $\exists M >0$, $n>M \implies |\frac{S_n}{n(n+1)}| \leq \frac14$
On $\{ \omega: |X_n| \geq n \text{ }i.o.\}= \{ \omega: \frac{|X_n|}{n} \geq 1 \text{ } i.o.\}$
\begin{align} \left| \frac{S_n}{n}-\frac{S_{n+1}}{n+1}\right| &= \left| \frac{S_n}{n(n+1)}-\frac{X_{n+1}}{n+1}\right| \\ & \geq \left|\left|\frac{X_{n+1}}{n+1} \right| - \left| \frac{S_n}{n(n+1)} \right|\right| \\ &\geq 1-\frac14 \text{ } i.o.\\ &> \frac23 \end{align}