# Reverse Stolz-Cesaro Theorem

One form of Stolz-Cesaro that is well known states:

If $(a_n)$ and $(b_n)$ are real sequences such that $b_n \uparrow \infty$ and $(a_{n+1} - a_n)/(b_{n+1} - b_n)$ converges as $n \to \infty$, then

$$\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L \in \mathbb{R} \cup\{\pm \infty\} \,\, \implies \,\, \lim_{n \to \infty}\frac{a_n}{b_n} =L$$

It is perhaps not so well known that a reverse implication is true when an additional condition is imposed:

$$\lim_{n \to \infty}\frac{b_n}{b_{n+1}} = B \in \mathbb{R} \setminus \{1\} , \,\,\,\,\, \lim_{n \to \infty}\frac{a_n}{b_n} = L \,\, \implies \,\,\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = L$$

For a proof note that

$$\frac{a_{n+1} - a_n}{b_{n+1} - b_n} \left(1 - \frac{b_n}{b_{n+1}} \right) = \frac{a_{n+1}}{b_{n+1}} - \frac{a_n}{b_{n+1}} = \frac{a_{n+1}}{b_{n+1}} - \frac{a_n}{b_{n}} \frac{b_n}{b_{n+1}}$$

and, hence,

$$\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \left(1 - \frac{b_n}{b_{n+1}} \right)^{-1} \left(\frac{a_{n+1}}{b_{n+1}} - \frac{a_n}{b_{n}} \frac{b_n}{b_{n+1}} \right)$$

If $B \neq 1$, the limit on the RHS exists and

$$\lim_{n \to \infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} = (1- B)^{-1}(L - L B) = L$$

If $B = 1$, then counterexamples to the reverse implication are readily found.

The question is what additional conditions are needed for the reverse Stolz -Cesaro theorem to hold in the case where $B = 1$?

In view of the counterexample $a_n=n+(-1)^n$, $b_n=n$, we see that $$b_n/b_{n+1} \rightarrow 1, \ \ a_n/b_n \rightarrow 1 \ \ \textrm{ as }n\rightarrow\infty.$$ This gives $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n} = \frac{1+(-1)^n(-1-1)}1$$ and it does not converge. Moreover, this example has the property that $$c_n=a_n-1 \cdot b_n = (-1)^n, \ \$$ then $c_{n+1}-c_n$ does not converge.
This suggests that we must have an additional condition that the sequence $$c_n = a_n - Lb_n$$ satisfies $$\frac{c_{n+1}-c_n}{b_{n+1}-b_n} \rightarrow 0 \ \textrm{ as } n\rightarrow \infty.$$