The fact that $b_n\to \infty$ is quite easy to check: if not, there is a $M$ and a subsequence $(b_{n'})$ which remains below $M$, hence $\frac 1{b_{n'}}\max_{1\leqslant j\leqslant n'}|X_j|\geqslant \frac 1{M}\max_{1\leqslant j\leqslant n'}|X_j|$, hence $\max_{1\leqslant j\leqslant n'}|X_j|$ would converge to $0$ in probability, which is not possible.

Let us denote the convergence in distribution by $\Rightarrow$.

Theorem 7.6 in Durrett's book *Probability: theory and examples*, (second edition), provides an useful theorem here:

**The convergence of types theorem.** Assume that $Y_n\Rightarrow Y$ and there are constants $\alpha_n>0$, $\beta_n$ such that $Y'_n=\alpha_nY_n+\beta_n\Rightarrow Y'$, where $Y$ and $Y'$ are not degenerate. Then there are $\alpha>0$ and $\beta\in\Bbb R$ such that $\alpha_n\to \alpha$ and $\beta_n\to \beta$.

The proof uses the fact that in case of convergence in distribution, the sequence of corresponding characteristic functions actually converges uniformly on compact sets. Then we proves that for $\alpha$, there is an unique positive real number which can be a candidate, and the same for $\beta_n$.

Here, we use this theorem with $W_n:=\frac{S_n-a_n}{b_n}$, $\alpha_n:=\frac{b_n}{b_{n+1}}$ and $\beta_n:=\frac{a_n-a_{n+1}}{b_{n+1}}$. It works since $\frac{X_{n+1}}{b_{n+1}}$ goes to $0$ in probability.

Finally, we have to use independence in order to identify the obtained limits. We compute the characteristic function of $W_n$, and of $W_{n+1}$, and we take the limits.