Stronger version of strong law of large numbers 
Let $(X_i)_{i\in\mathbb{N}}$ be pairwise independent random variables where $E\left[X_i\right]=0$ for all $i\in\mathbb{N}$ and $\sup_{n}E\left[X_n^2\right]\lt\infty$. Then for $S_n=\sum_{i=1}^n X_i$ and $t>\dfrac{3}{4}$ we have $\dfrac{S_n}{n^t}\to 0$ almost surely.

I looked at different proofs of the SLLN, but I can't see where we could be using such a $t$. Given the assumptions I was able to see we have finite variance, but I didn't make much ground in that direction. Can anyone give me any hints to get me started?
 A: It seems that the wanted result can be improved in two directions. 


*

*The random variables do not need to be pairwise independent: uncorrelatedness is sufficient.

*We have the result for $t\gt 1 /2$ and not only $3/4$. 


Using the fact that the random variables are uncorrelated, we derive easily that 
$$\mathbb E\left[S_n^2\right]\leqslant n\sup_{i\geqslant 1}\mathbb E \left[X_i^2\right].$$
Now, we shall show the following:

Lemma. Let $\left(X_i\right)_{i\geqslant 1}$ be a centered uncorrelated sequence such that $\sup_{i\geqslant 1}\mathbb E \left[X_i^2\right]$ is finite. Then there exists a constant $C$ such that for any $n\geqslant 1$, $$\sup_{n\geqslant 1}\frac 1{2^nn^2}     \mathbb E \left[\max_{1\leqslant j\leqslant 2^n}     S_{j}  ^2\right]\leqslant C.$$
  This is contained in a paper by Stout (1974). 

Let $s:=\sup_{i\geqslant 1}\mathbb E\left[X_i^2\right]$.             We shall show by induction on $n$ that for any uncorrelated sequence $\left(X_i\right)$ with partial sums $S_n$,
$$\frac 1{2^nn^2}     \mathbb E \left[\max_{1\leqslant j\leqslant 2^n}     S_{j}  ^2\right]\leqslant \max\left\{2s          ,  2\mathbb E\left[\max\left\{X_1^2,S_2^2\right\}\right]     \right\} =:A       .$$
This is satisfied for $n=1$ by assumption. Now, assume that it is true for some $n$. Denote $M_n:=\max_{1\leqslant j\leqslant 2^n}\left\lvert S_j\right\rvert$. Then 
$$M_{n+1}^2\leqslant M_n^2+\max_{1\leqslant j\leqslant 2^n}\left\lvert S_{2^n+j}-S_{2^n}+S_{2^n}\right\rvert^2                     $$
hence 
$$\mathbb E\left[M_{n+1}^2 \right] \leqslant \mathbb E\left[M_{n}^2 \right]
+\left( \left\lVert \max_{1\leqslant j\leqslant 2^n}\left\lvert S_{2^n+j}-S_{2^n}\right\rvert\right\rVert  _2 +\left\lVert S_{2^n}\right\rVert_2      \right)^2. $$
Using the induction assumption, we obtain that $\mathbb E\left[M_{n}^2 \right]\leqslant n^22^nA$; using again the induction assumption with the sequence $\left(X_{i+2^n}\right)_{i\geqslant 0}$ instead of the original sequence, we obtain that  $\left\lVert \max_{1\leqslant j\leqslant 2^n}\left\lvert S_{2^n+j}-S_{2^n}\right\rvert\right\rVert  _2^2\leqslant n^22^nA$. Consequently, 
$$\mathbb E\left[M_{n+1}^2 \right] \leqslant n^22^nA +\left( \sqrt{  n^22^nA}   +2^{n/2}s^{1/2}     \right)^2=2^{n+1}n^2A+2^{n+1}\sqrt As^{1/2}  +2^ns.$$
We have to show that
$$2^{n+1}n^2A+2^{n+1}\sqrt As^{1/2}  +2^ns\leqslant 2^{n+1}\left(n+1\right)^2A     $$
which is equivalent to 
$$2\sqrt As^{1/2}  +s\leqslant 2 A\left(2n+1\right).$$
It is thus sufficient to prove that $2\sqrt As^{1/2}  +s\leqslant 2 A$, which holds since $A\geqslant 2s$.
Now, to conclude, observe that for any $t\gt 1/2$, the expectation 
of 
$$\sum_{n=1}^N2^{-2nt}\max_{1\leqslant j\leqslant 2^n}\left\lvert S_i\right\rvert^2               $$ can be bounded independently of $N$, hence the random variable $\sum_{n=1}^{+\infty}   2^{-2nt}\max_{1\leqslant j\leqslant 2^n}\left\lvert S_i\right\rvert^2$ is almost surely finite hence $2^{-nt}\max_{1\leqslant j\leqslant 2^n}\left\lvert S_i\right\rvert$ goes to zero almost surely.
