Reverse Stolz-Cesaro Theorem

This might be too artificial, but I think it is required to do this.

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 $$ \frac{c_{n+1}-c_n}{b_{n+1}-b_n} \rightarrow 0 \ \textrm{ as } n\rightarrow \infty. $$