Kolmogorov proved, that, as one considers independent (not necessary equally distributed) Random Variables: $\{X_n\}_{n\ge0}\subseteq \mathcal L^2$
With $\mathrm{Var} (X_n)=\sigma^2_n$ and without loss of generality $E[X_n]=0$.
If $\sum_{n=0}^\infty \frac{\sigma^2_n}{n^2} \lt \infty$ then SLLN holds, that is:
$$\frac1n\sum_{k=0}^n X_k \rightarrow 0 \,\,\,\,\,\,\,\,\,\,\,\,\,\text{a.s.}$$
Something different:
Also it is known, that if $\sup_{n\ge0}\sigma^2_n =: v \lt \infty$ pairwise uncorrelated $(X_n)_{n\ge0}$ will suffice for SLLN to hold.
My question in view of this situation: Is it possible to weaken the condition of $(X_n)_{n\ge0}$ beeing independent and instead merely consider pairwise uncorrelated $(X_n)_{n\ge0}$?
If you want to impose just uncorrelatedness between the variables in the sequence rather than independence, then you need to impose a stronger condition than $\sum_{i=1}^{\infty} \operatorname{Var}[X_i]/i^2 < \infty$. To be more precise, a sufficient condition in this case that will guarantee SLLN is the following: $$\sum_{i=1}^{\infty} \operatorname{Var}[X_i] \left(\frac{\log i}{i}\right)^2 < \infty.$$
This follows from Serlfing's SLLN.