For any i.i.d. $X_n$ it follows $\lim_{n\to\infty}X_n/n=0$ pointwise? Let $X_n$ denote a sequence of i.i.d. random variables and $Y_n=X_n/n$.
$Y_n$ converges pointwise towards Y if
$~~~{\displaystyle \lim _{n\to \infty }Y_{n}(\omega )=Y(\omega ),\,\,\forall \omega \in \Omega}$,
where $\Omega$ is the sample space of the underlying probability space over which the random variables are defined.
It seems trivial that $\lim_{n\to\infty}X_n(\omega)/n=0$, $\forall \omega \in \Omega$ and i.i.d. distributions of $X_n$.
Then, also all other kinds of convergence are true. What do I miss?
I'm confused by the comments of experienced guys in this post:
Proving almost sure convergence
 A: Here is an example for iid variables where $X_n(\omega)/n$ does not converge to zero.
Suppose that $X_n$ are iid copies of the St. Petersburg random variable. That is, you repeatedly flip a coin until the result is heads, and let $X=2^{\text{# flips}}$.
I claim that with probability one, $X_n(\omega)/n$ will not converge to zero. To show this, I will show that $X_n(\omega)/n\ge 1$ will occur infinitely often with a probability of one.
For any $n$ in the range $2^k$ to $2^{k+1}-1$, we have
$$
P(X_n/n\ge 1)=P(X_n\ge n)= P(X_n\ge 2^k)=P(\hbox{# flips}\ge k)=2^{-(k-1)}
$$
Therefore, the probability that all of the numbers $X_n/n$ in the range $2^k$ to $2^{k+1}-1$ are less than one is at most
$$
P\left(\bigcap_{n=2^k}^{2^{k+1}-1}\{X_n/n<1\}\right)\le (1-2^{-(k-1)})^{2^k}=\left(1-\frac{2}{2^k}\right)^{2^k}\approx e^{-2}.
$$
so the probability that at least one of the $X_n/n\ge 1$ occurs is $1-e^{-2}\approx 86\%$. Therefore, in each of then intervals $[2^k,2^{k+1})$, there is at least an $86\%$ chance that there will be a variable $X_n/n\ge 1$. This these events are all independent, in the long run, it will occur infinitely many times.

It turns out that the same thing will happen whenever $X$ has infinite expectation. The intuition is that having a large expectation increases the chance of $X$ being large, so it turns out infinite expectation implies $X_n$ will actually be larger than $n$ infinitely often.
