Strong and Weak Law of Large Numbers Let $(X_n)$ be a sequence of independently Poisson distributed random variables with parameter $\lambda = n+1$ ($n=1,2,...$). I want to check whether SLLN holds for $Y_n = \frac{X_n}{\sqrt {\log(n+1)}}$.
Well, it's pretty easy to prove that WLLN holds but it's hard for me to prove or disprove SLLN. I tried to use the Borel-Cantelli lemma but with no result.
 A: Here is a partial argument that shows probability 1 convergence over a sparse subsequence. 
As in my comment above, it seems best to use a 4th centralized moment: For any $\epsilon>0$ we have
\begin{align}
P\left[\left|\frac{1}{n}\sum_{i=1}^n (Y_i-E[Y_i])\right|\geq \epsilon\right]&\leq \frac{E\left[\left(\sum_{i=1}^n (Y_i-E[Y_i]) \right)^4\right]}{\epsilon^4n^4} \\
&= \frac{1}{\epsilon^4n^4}\sum_{i=1}^n E[(Y_i-E[Y_i])^4] \\
&\quad + \frac{1}{\epsilon^4n^4}\sum_{i=1}^n\sum_{j\in\{1, ..., n\} - i}E[(Y_i-E[Y_i])^2]E[(Y_j-E[Y_j])^2]\\
&\leq \frac{1}{\epsilon^4n^4}\sum_{i=1}^n E[(Y_i-E[Y_i])^4] \\
&\quad + \frac{1}{\epsilon^4n^4}\sum_{i=1}^n\sum_{j=1}^nE[(Y_i-E[Y_i])^2]E[(Y_j-E[Y_j])^2]\\
\end{align}
The second term is the dominant term and so by the bound that you gave in your comments:
$$ \frac{1}{\epsilon^4}\left(\frac{1}{n^2}\sum_{i=1}^nE[(Y_i-E[Y_i])^2]\right)\left(\frac{1}{n^2}\sum_{j=1}^nE[(Y_j-E[Y_j])^2]\right) \leq \frac{C}{\epsilon^4(\log(n+1))^2}$$ 
where $C$ is some positive constant, 
which indeed gives us the $1/(\log(\cdot))^2$ as we wanted. Now you can fix $\delta>0$ and sample at the sparse subsequence of times $n_k = \lceil(1+\delta)^k\rceil$ to get probability 1 convergence over that sparse subsequence of times: 
$$\lim_{k\rightarrow\infty} \frac{1}{n_k}\sum_{i=1}^{n_k}(Y_i-E[Y_i]) = 0 \quad \mbox{ with prob 1} $$

Edit: My original posted answer concluded too hastily that "standard nonnegativity arguments" imply convergence over the sparse subseqeunce $\{n_k\}_{k=1}^{\infty}$ implies convergence over the full sequence $\{n\}_{n=1}^{\infty}$. However, while the $Y_i$ variables are nonnegative, their time average is not converging to a finite mean $m$, so it is not clear how to proceed by "standard" ways.  
