Almost sure convergence of equal weighted sum Let $Z_1, Z_2, ...$ be independent random variables in the same probability space defined as follows:

$$P(Z_n=n)=P(Z_n=-n)=\frac{1}{2n^2} \space \mathrm{and} \space P(Z_n=0)=1-\frac{1}{n^2}$$

Is it true that

$$\lim \frac{1}{\sqrt{n}}(Z_1+...+Z_n)  = 0 \ \text{a.s.}$$

?

I thought of using the Borel-Cantelli Lemmas, but I don't know how to calculate the probability that $$P(|Z_1 + \ldots + Z_n| \ge \epsilon)$$ where $\epsilon &gt; 0$.

I already tried to apply Chebyshev's inequalities, but it didn't work.
Notice that the series $\sum_{n=1}^{+\infty}\mathbb P\left(Z_n\neq 0\right)$ converges, as $\mathbb P\left(Z_n\neq 0\right)=1/n^2$. This means that for almost every $\omega\in \Omega$, we can find an integer $N\left(\omega\right)$ such that if $n\geqslant N\left(\omega\right)$, then $Z_n\left(\omega\right)  = 0$. This implies that for $n\geqslant N\left(\omega\right)$, 
$$\frac{1}{\sqrt{n}}\left(Z_1\left(\omega\right)+\dots+Z_n\left(\omega\right)\right)=\frac{Z_1\left(\omega\right) +\dots+Z_{N\left(\omega\right)}\left(\omega\right)}{\sqrt n}    ,$$ 
which gives the wanted result. 

Two remarks:


We did not use independence of the sequence $\left(Z_n\right)_{n\geqslant 1}$.
The normalization $\sqrt n$ could be replaced by $a_n$ where $a_n\to +\infty$ as $n$ goes to infinity.
Hint:

Using BCL1, it looks like we have

$$P(\liminf(Z_n \ne n)) = 1$$

$$P(\liminf(Z_n \ne -n)) = 1$$

I think we can conclude that

$$P(\liminf(Z_n = 0)) = 1 \tag{*}$$

because:

$\exists m_1 \ge 1$ s.t. $Z_{m_1} \ne m_1$, $Z_{m_1+1} \ne m_1+1$, ...

$\exists m_2 \ge 1$ s.t. $Z_{m_2} \ne m_2$, $Z_{m_2+1} \ne m_2+1$, ...

So I think for $m := \max\{m_1, m_2\}$, we have

$$Z_m \ne m, -m$$

$$Z_{m+1} \ne m+1, -(m+1)$$

$$\vdots$$

which I think is $(*)$



Also not sure if relevant but by BCL2, it looks like

$$P(\limsup(Z_n = 0)) = 1$$



As for computing

$$p_n := P(|\sum_{i=1}^{n} Z_i| &gt; 0)$$

If you're going to use BCL2 on $\sum p_n$, assuming it converges, it looks like you'll be able to conclude

$$P(\limsup(|\sum_{i=1}^{n} Z_i| &gt; 0)) = 0$$

$$\to P(\liminf(|\sum_{i=1}^{n} Z_i| = 0)) = 1$$

$$\to P(\liminf(\sum_{i=1}^{n} Z_i = 0)) = 1$$

This looks pretty strong to me, but I guess it'll work assuming you compute $p_n$ and $\sum p_n$ converges.