This is exercise 10 page 138 in Chung's book: a course in probability theory. Let $(X_n)_n$ be a sequence of independent random variables, $Y_n=\sum_{k=1}^nX_k$, $\alpha \in ]0;2[,\alpha \neq 1,$ and two constants $c_1$ and $c_2$ such that: $$\forall n, \forall x>0,\frac{c_1}{x^\alpha} \leq P(|X_n|>x) \leq \frac{c_2}{x^\alpha}.$$ If $\alpha>1,$ suppose also that $E[X_n]=0$ for all $n$. Then for any sequence $(u_n)_n$ increasing to infinity, prove that:
$P(|Y_n|>u_n \ i.o.)=0$ or $1$ if $\sum_n \frac{1}{u_n^\alpha}< \ or \ =\infty$,
If we suppose that $\sum_n \frac{1}{u_n^\alpha}<\infty,$ then we have $\sum_nP(|X_n|>u_n) \leq c_2\sum_n \frac{1}{u_n^\alpha}<\infty$
if $\alpha<1$ then $\sum_n\frac{1}{u_n}E[|X_n|1_{|X_n| \leq u_n}]\leq c_2\sum_n\frac{1}{u_n}\int_0 ^{u_n}\frac{1}{x^\alpha}dx<\infty.$
if $\alpha>1,$ then $\sum_n\frac{1}{u_n}|E[X_n1_{X_n>u_n}]| \leq c_2\sum_n \int_0^{\infty}\frac{1}{(\max(x,u_n))^\alpha}dx \leq 2c_1 \sum_n \frac{1}{u_n^{\alpha}}<\infty$.
Using that same way, we can prove that $\sum_n {Var(\frac{1}{u_n}X_n1_{X_n \leq u_n})}<\infty$
By Kolmogorov three series theorem, $\sum_n\frac{X_n}{u_n}$ converges a.s and Kronecker's lemma gives us: $\frac{1}{u_n}\sum_n X_n$ converges to 0 a.s, which implies $P(|Y_n|>u_n \ i.o.)=0.$
If we suppose that $\sum_n\frac{1}{u_n^\alpha}=\infty,$ we can prove that $\sum_nP(|X_n|>u_n)=\infty,$ then $\sum_n\frac{X_n}{u_n}$ diverges a.s but I don't know how to deduce that $P(|Y_n|>u_n \ i.o)=1.$
I appreciate if you have any hints to conclude.