I'm working my way through Rick Durrett's book "Probability: Theory and Examples" and want help solving the following question.
$\textbf{Question:}$ Suppose $X_1,X_2,\ldots$ are i.i.d random variables, $\mathbb{P}(0\leqslant X_i<\infty)=1$ and $\mathbb{P}(X_i>x)>0~\forall~x.$ Let, $\mu(s)=\int\limits_{0}^s x\mathrm{d} F(x)$ and $\nu(s)=\frac{\mu(s)}{s(1-F(s))}$. It is known that there exist constants $a_n$ so that $\frac{S_n}{a_n}\to1$ in probability, iff $\nu(s)\to\infty$ as $s\to\infty$. Pick $b_n\geqslant 1$ so that $n\mu(b_n)=b_n$ (this works for large $n$), and use weak law for triangular arrays (theorem 2.2.6 in the book) to prove that the condition is sufficient.
$\textbf{Attempt:}$ Theorem 2.2.6 says if as $n\to\infty$ the conditions (i) $\sum\limits_{k=1}^{n}\mathbb{P}(|X_{n,k}|>b_n)\to 0$ and (ii) $\sum\limits_{k=1}^{n}\mathbb{E}[\bar{X}_{n,k}^2]\to0$ holds and if for $S_n=\sum\limits_{k=1}^{n}X_{n,k}$ and $a_n=\sum\limits_{k=1}^{n}\mathbb{E}[\bar{X}_{n,k}]$ then $\frac{S_n-a_n}{b_n}\to 0$ in probability.
One of the first things we must do is select $b_n$. I fail to find any motivation or intuition to select an appropriate $b_n$. Secondly, I could do the following calculation for large $n$. Since $n\mu(b_n)=b_n$ as $n$ becomes large, we can do the following: $$ \mathbb{P}(|X_n|>b_n) = 1-\mathbb{P}(|X_n|\leqslant b_n) = 1-F(b_n)\Rightarrow n\mathbb{P}(|X_n|>b_n) = \frac{b_n}{\mu(b_n)}(1-F(b_n))=\frac{1}{\nu(b_n)}. $$ However, I don't know what to do next and how to proceed further.