First, recall that Kolmogorov's Three Series Theorem (K3ST) is used to prove that the
sum of independent random variables $\sum_{i=1}^{\infty} S_i$ converges a.s. ("a.s." means "almost surely"), i. e., $S_1+\cdots +S_n$ converges a.s. But for this problem, you are asked to prove that $n^{-1/\alpha} (X_1 + \cdots + X_n)\to 0$ a.s. At first glance, it seems like the conclusion of K3ST looks nothing like what we want to prove and so won't be useful. Kronecker's Lemma (http://en.wikipedia.org/wiki/Kronecker%27s_lemma) will provide the required bridge from K3ST to the conclusion we wish to prove.
Define the random variable $S_n$ to be
$$
S_n = \frac{1}{n^{\frac{1}{\alpha}}} X_n.
$$
We will check the three conditions of K3ST for $S_n$. This will allow us to conclude $S_1+\cdots+S_n$ converges a.s. Simple application of Kronecker's Lemma with $S_n$ and the constant sequence $n^{\frac{1}{\alpha}}$ will then allow us to conclude $n^{-1/\alpha} (X_1 + \cdots + X_n)\to 0$ a.s.
For brevity, define the random variable $C_n$ as
$$
C_n = \lceil|X_n|^{\alpha}\rceil.
$$
Also, define
$$
p_i = \mathbb{P}(C_n = i)
$$
for all integers $i\ge 0$. Note that $p_i$ does not depend on $n$, since $X_n$ and therefore $S_n$ and $C_n$ are i.i.d.
By Minkowski's inequality and the fact that $C_n\le |X_n|^{\alpha} + 1$, we conclude $\mathbb{E}(C_n) < \infty$.
We now verify K3ST for $S_n$.
Series 1 We must verify that $\sum_{n\ge 1} \mathbb{P}(|S_n| > 1) < +\infty$. Observe
\begin{eqnarray}
\sum_{n\ge 1} \mathbb{P}(|S_n| > 1) &\le& \sum_{n\ge 1} \mathbb{P} (|X_n| > n^{\frac{1}{\alpha}}) \\
&=& \sum_{n\ge 1} \mathbb{P} (|X_n|^{\alpha} > n) \\
&\le& \sum_{n\ge 1} \mathbb{P} (C_n > n) \\
&=& \sum_{n\ge 1} \sum_{i=n+1}^{\infty} p_i \\
&=& \sum_{i\ge 2} (i-1) p_i\\
&\le& \mathbb{E}(C_n) < \infty.
\end{eqnarray}
Series 2 We need to show $\sum_{n\ge 1} \mathbb{E}(S_n 1_{|S_n|\le 1})$ converges. One can show this by proving
$$
(*)\ \ \ \ \ \ \sum_{n\ge 1} \mathbb{E}(|S_n| 1_{|S_n|\le 1}) < \infty.
$$
To show this, first note
\begin{eqnarray}
\mathbb{E}(|X_n| 1_{|X_n|\le n^{1/\alpha}}) &=& \mathbb{E}(|X_n| 1_{|X_n|^{\alpha}\le n}) \\
&\le& \mathbb{E}(C_n^{1/\alpha} 1_{C_n\le n}) \\
&=& \sum_{i=1}^n i^{\frac{1}{\alpha}} p_i.
\end{eqnarray}
Therefore,
\begin{eqnarray}
\sum_{n\ge 1} \mathbb{E}(|S_n| 1_{|S_n|\le 1}) &=& \sum_{n\ge 1} \frac{1}{n^{\frac{1}{\alpha}}} \mathbb{E}(|X_n| 1_{|X_n|\le n^{1/\alpha}}) \\
&\le& \sum_{n\ge 1} \frac{1}{n^{\frac{1}{\alpha}}} \sum_{i=1}^n i^{\frac{1}{\alpha}} p_i \\
&=& \sum_{i\ge 1} \left(\sum_{n\ge i} \frac{1}{n^{\frac{1}{\alpha}}}\right) \ i^{\frac{1}{\alpha}} p_i \\
&\le& K \sum_{i\ge 1} i p_i = K \mathbb{E}(C_n) < \infty.
\end{eqnarray}
Here, $K$ is a finite constant, independent of $i$, such that
$$
\sum_{n\ge i} \frac{1}{n^{\frac{1}{\alpha}}} \le \frac{K}{i^{\frac{1}{\alpha} - 1}}.
$$
Series 3 We must show $\sum_{n\ge 1} \mathrm{Var}(S_n 1_{|S_n|\le 1}) < \infty$. Given $(*)$ above, this is easy:
\begin{eqnarray}
\sum_{n\ge 1} \mathrm{Var}(S_n 1_{|S_n|\le 1}) &\le& \sum_{n\ge 1} \mathbb{E}(S^2_n 1_{|S_n|\le 1}) \\
&\le& \sum_{n\ge 1} \mathbb{E}(|S_n| 1_{|S_n|\le 1}) < \infty.
\end{eqnarray}
And we're done.
