Exercise of converges to type Suppose $\left\{X_n , 1\le n\right\}$ are i.i.d. random variables with common distribution $F$
and set $S_n = \sum_{i=1}^n X_i$. Assume that there exist $a_n>0 , b_n \in R$ such that 
$$\frac{S_n}{a_n}-b_n\Rightarrow Y$$
where $Y$ has a non-degenerate proper distribution. Use the convergence to types theorem to show that 
$$a_n\rightarrow \infty , \frac{a_n}{a_{n+1}}\rightarrow1$$
(Symmetrize to remove $b_n$. You may want to first consider ${a_{2n}}/{a_n}$).
This problem is an exercise 8.8.26 from A Probability path, Resnick.
I checked that ${a_{2n}}/{a_n}$ converges to some $a>1$ but I can't show that $a=\infty$.
How to prove it?
 A: I will assume that the $X_n$ are symmetric and $b_n=0$. Notice that no subsequence of $\left(a_n\right)_{n\geqslant 1}$ converge to $0$. Otherwise, we would have that $a_{n_j}\to 0$, and if we choose $R$ a continuity point of the cumulative distribution function of $\left|Y\right|$ and $\delta \gt 0$, then we have $a_n\lt \delta/R$ for $n$ large enough hence
$$\Pr\left( \frac{\left|S_n\right|}{a_n} \gt R\right) \geqslant \Pr 
\left(\left|S_n\right| \gt \delta\right).$$
This proves that $\limsup_{n\to +\infty}\Pr 
\left(\left|S_n\right| \gt \delta\right) \leqslant \Pr\left(\left|Y\right|>R\right)$. Since $R$ and $\delta$ are arbitrary and $Y$ has a proper distribution, we would get that $S_n\to 0$ in probability, but $X_1$ has a non degenerated distribution, which gives a contradiction. 
It suffices to prove that $a_n\to \infty$; once it is done, we notice that 
$$\frac{S_{ n+1}}{ a_{n+1} }=\frac{S_n}{a_n}   \frac{a_n}{a_{n+1} }  +\underbrace{ \frac{X_{n+1}}{a_{n+1}} }_{ \Rightarrow 0 }                  $$
and the convergence of types shows that the sequence $\left(a_{n+1}/a_n\right)_{n\geqslant 1}$ is convergent, say to some $\ell$. We thus have $Y=\ell Y$ in distribution, and since $Y$ is not degenerated. 
Assume that the sequence $\left(a_n\right)_{n\geqslant 1}$ does not go to infinity. Then there exists some positive $R$  and an increasing sequence of integers $\left(n_j\right)_{j\geqslant 1}$ such that $0\lt a_{n_j}\leqslant R$. We extract a further converging subsequence, denoted in the same way for simplicity. By the initial remark, the limit is necessarily some positive $\ell$. We thus deduce that $\left(S_n /\ell\right)_{n\geqslant 1}$ converges to some non-degenerated limit, a contradiction.      
