# Converse of Stolz-Cesaro

It is not difficult to prove a converse to Stolz-Cesaro in the form:

If $a_n$ and $b_n$ are sequences satisfying:

a) $b_n$'s are increasing and divergent

b)$\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}}=l\in \mathbb{R}$

c)$\lim_{n\rightarrow \infty}\frac{b_{n}}{b_{n+1}}=L \in \mathbb{R}-\{1\}$

Then we have that: $$\lim_{n\rightarrow \infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=l$$

I am interested in the case when the $L$ above equals to 1. What sort of additional conditions can we put (on the series $b_n$ ) such that we have a converse to Stolz-Cesaro theorem (similar to above)?

Thanks

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If the limit $b_n/b_{n+1}$ is $1$, you expect plenty of problems ; first of all, you know that the denominator $b_{n+1} - b_n$ will be ill-behaved, because in the case where $L \neq 1$ you expect $b_{n+1} - b_n \sim (L-1) b_n$, which goes to $+\infty$ when $L > 1$ and $-\infty$ when $L < 1$. However, when $L=1$, you cannot predict its behavior. – Patrick Da Silva Jul 21 '13 at 8:04