The Converse Stolz-Cesaro theorem states that if $(b_n)_n$ is strictly monotone and divergent and:
$$ \lim_{n \to \infty}\frac{b_n}{b_{n+1}} = B \in \mathbb{R} \setminus \{1\} , \,\,\,\,\, \lim_{n \to \infty}\frac{a_n}{b_n} = L\in\mathbb{R}\cup\{\pm\infty\} \,\, \implies \,\,\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L $$
In every book/article/internet post I found the following proof which is clearly true if $L\in\mathbb{R}$:
$$\frac{a_{n+1} - a_n}{b_{n+1} - b_n} \left(1 - \frac{b_n}{b_{n+1}} \right) = \frac{a_{n+1}}{b_{n+1}} - \frac{a_n}{b_{n+1}} = \frac{a_{n+1}}{b_{n+1}} - \frac{a_n}{b_{n}} \frac{b_n}{b_{n+1}}$$
and, hence,
$$\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \left(1 - \frac{b_n}{b_{n+1}} \right)^{-1} \left(\frac{a_{n+1}}{b_{n+1}} - \frac{a_n}{b_{n}} \frac{b_n}{b_{n+1}} \right)$$
Since $B \neq 1$, the limit on the RHS exists and
$$\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = (1- B)^{-1}(L - L B) = L$$
My question is: What if $L=+\infty$? Does the same result hold? Because the last relation might be false for $B>0$...