Let $X_k$ be a sequence of IID random variable with $E(X_k)=0$ and finite variance. Show that $\frac{S_n}{n^p} \xrightarrow{a.s.} 0$ for p > 1/2 My idea to this question is
$\frac{S_n}{n^p}=\frac{S_n}{\sigma\sqrt{n}}\cdot\frac{\sigma}{n^{p-1/2}}$
By CLT,
$\frac{S_n}{\sigma\sqrt{n}} \xrightarrow{d} N(0,1)$
Also, it can be shown
$\frac{\sigma}{n^{p-1/2}} \xrightarrow{d} 0$
Therefore,
$\frac{S_n}{\sigma\sqrt{n}}\cdot\frac{\sigma}{n^{p-1/2}}=\frac{S_n}{n^p} \xrightarrow{d} 0$
Since it converges to a constant, it implies
$\frac{S_n}{n^p} \xrightarrow{p} 0$
We know if $\frac{S_n}{n^p} \xrightarrow{p} 0$, then there is a subsequence $\frac{S_{n_{k}}}{n_k^p} \xrightarrow{a.s.} 0$.
My question is how to show such subsequence $\frac{S_{n_{k}}}{n_k^p}\xrightarrow{}  \frac{S_n}{n^p}$ ?
 A: Hmmm, it's very weird that some computational trick doesn't work on this. So let's do it this way.
Demonstration
For simplicity, I'll call out a function $h$ from $\mathbb{N}$ taking values in $\mathbb{N}$ with some specific properties without explicitly specifying it.
Let $h$ be a function from $\mathbb{N}$ to $\mathbb{N}$ such that:

*

*$h$ is strictly increasing, thus $h(n) \ge n$

*$ \sum_{n \ge 1} \dfrac{1}{ h(n)^{2p-1}} < +\infty$

*$  \dfrac{h(n)}{h(n+1)} > a$ for all $n$ for some $a$ positive

(A candidate for $h$ is $h(n)=n^q$ for $q$ big enough)
Also, for simplicity, WLOG: $\sigma =1$
By Doob's, we have
$$ \mathbb{E}\left( \max_{  m \le h(n)} |S_{m}|^2 \right) \le 4\mathbb{E}( (S_{h(n)})^2 )= 4h(n)$$
Thus,
$$ \sum_{n \ge 1} \mathbb{E}\left( \dfrac{1}{h(n)^{2p}}\left( \max_{  m \le h(n)} |S_{m}|^2 \right) \right) \le \sum_{n \ge 1} \dfrac{4}{h(n)^{2p-1}} < +\infty$$
Thus,
$$ \lim_{n \rightarrow +\infty} \dfrac{1}{h(n)^{p}}\max_{   m \le h(n)} |S_{m}| = 0 \quad a.e \quad (**)$$
As $h(m)> a h(m+1)$, we have that for all $n$:
$$  \dfrac{1}{n^{p}} |S_n| \le \dfrac{1}{h(g(n)-1)^p} \max_{   k \le h(g(n))}  |S_{k}| \le \dfrac{1}{a^p} \dfrac{1}{ h(g(n))^p} \max_{   k \le h(g(n))}  |S_{k}| $$
Where $g(n)$ is defined as the smallest number such that $ h(g(n)-1 ) \le m \le h(g(n))$.
As $h$ increases to infinity, $g$ must also increase to infinity.
Hence, conclusion $\square$
Comments

*

*The same method can be applied to prove that $\dfrac{S_n}{ n^{1/2}\ln(n)^s } \longrightarrow 0 \quad a.e$ for all $s>1/2$


*If $(X_i)$ are gaussian, we can even have a tight bound for $\limsup |S_n|$


*Many conditions in the statement can be relaxed.


*If I remember correctly, the rate when $(X_i)$ are Gaussian is $O( \dfrac{1}{ n^{1/2} \sqrt{ ln(ln(n))} } )$
