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 $B>0$…
Best Answer
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...