Stolz-Cesaro converse theorem proof - possible mistake? 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 $B>0$...
 A: This theorem seems to fail for $L=+\infty$. Here is an example. Please let me know if there is anything wrong.
Let $b_n=2^n$ and $a_{2k}=a_{2k-1}=k\cdot 2^{2k}$ for $k\geq 1$, then
$$\frac{a_{2k-1}}{b_{2k-1}}=\frac{a_{2k-1}}{2^{2k-1}}=2k\to+\infty,\qquad \frac{a_{2k}}{b_{2k}}=\frac{a_{2k}}{2^{2k}}=k\to+\infty,$$
hence $\lim_{n \to \infty}\frac{a_n}{b_n} =+\infty$. However,
$$\frac{a_{2k} - a_{2k-1}}{b_{2k} - b_{2k-1}}=0,\qquad \frac{a_{2k+1} - a_{2k}}{b_{2k+1} - b_{2k}}=\frac{(k+1)2^{2k+2}-k\cdot2^{2k}}{2^{2k}}\to+\infty.$$
Therefore, the limit $\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}$ does not exist.
However, here is a post by user @RRL stating the same theorem which includes the case $L=+\infty$. I'm not so confident about my example...
A: Let $r_n = a_n/b_n, s_n = b_n/b_{n+1}.$ The question becomes if we can design $r_n,s_n$ such that $r_n \to \infty, s_n \to B \neq 1, B \in \mathbb{R},$ but for $$ C_n := \frac{r_{n+1} - r_ns_n }{(1- s_n)},$$ it holds that $\liminf C_n < \infty$ - note that $\limsup C_n = \infty$ follows already from the standard Stolz-Cesaro argument.
The denominator of $C_n$ converges, so the issue just becomes that of controlling the numerator. But we can make the numerator dance up and down as we like by making $r_n$ dance up and down. For instance, take $s_n =1/2$ identically, and take $r_n$ that alternately jumps up by a factor of $4$ and down by a factor of $3$, i.e. fix $r_1 = 1$ and take $$ r_n = \begin{cases} 4r_{n-1} & n \textrm{ even}\\ r_{n-1}/3 & n \textrm{ odd}\end{cases}.$$
Note that $r_n = 4r_{n-2}/3$ and so $r_n \to \infty,$ but if $n$ is even then $r_{n+1} - r_n s_n = -r_n/6,$ so $\liminf C_n = -\infty$.
We can extend this to other values of $B$ - for $B \neq 1, B > 0,$ the above works by making the jump up factor larger than $B$ and the jump down smaller than $B$ (and ensuring that the overall 2-step jump up is $>1$).
If $B = 0$ then we can still have $\liminf C_n < \infty$ - for instance, jump down by $s_n/2,$ and give enough of a jump up to ensure that $r_n s_n \to \infty$ (e.g. jump up by $1/s_n^2$ type factors).
If $B < 0,$ then $C_n$ does diverge to $+\infty$ - for large $n$, $r_{n+1} - r_n s_n \ge r_{n+1} + r_n|B|/2.$
