Stolz-Cesaro theorem for 0/0 case when limit exists = 0

Consider this particular instance for the $$0/0$$ case of the Stolz-Cesaro Theorem: $$\lim_{n \to \infty} a_n = \lim_{n \to \infty} b_n = 0$$ with $$a_n$$ and $$b_n$$ two sequences of real numbers, $$b_n$$ strictly monotone, whereby $$\lim_{n \to \infty} \frac{a_{n+1} - a_{n}}{b_{n+1} - b_{n}} = 0 \,\,\,\,\,\,\,\, (1)$$ then also $$\lim_{n \to \infty} \frac{a_n}{b_n} = 0 \,\,\,\, \,\,\,\,(2)$$ it would appear to me that for those particular instances for which it is in addition verified that each $$b_n$$ is strictly positive and such that $$b_n \neq b_{n+1}$$, then the condition for the sequence $$b_n$$ to be strictly monotone could be dispensed with. Is this true, or am I missing some subtle features?

the counterexample shows that the conditions $$b_n$$ strictly positive, $$b_n \neq b_{n+1}$$, and limit (1) vanishing, are alone not sufficient to ensure that also limit (2) consequently vanishes. Were the sequence $$b_n$$ strictly monotone, then that would be true. Thus, even in such particular instances it would appear that no narrowing of the Stolz-Cesaro theorem hypotheses is feasible ... or, is it really so? For the instances whereby $$b_n \neq b_{n+1}$$, and limit (1) vanishes, instead of the usual "monotonicity" hypothesis it appears to me that it might suffice to verify this other additional condition: $$b_{n+1} \in O (b_{n}) \,\,\,\, AND \,\,\,\, b_{n+1} \in \Omega (b_{n})$$ and then also limit (2) will vanish. For a simple example let us take ($$\epsilon > 0$$): $$a_n = \frac{1}{n} \,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\, \,\,\,\,$$ $$b_n = \frac{1}{\sqrt{n}}\,\,\,\,\,\,\,\, @ \,\,\, even \,\,\, n$$
$$b_n = \frac{\epsilon}{\sqrt{n}}\,\,\,\,\,\,\,\, @ \,\,\, odd \,\,\, n$$ the $$b_n$$ sequence is not monotonic (except when $$\epsilon = 1$$, of course) but it satisfies both the Big Oh and Big Omega conditions (i.e. a Big Theta condition). And indeed, the vanishing of limit (1) implies also the vanishing of limit (2).

No, the monotonicity of $$(b_n)$$ is essential. If you investigate the proof of the Stolz-Cesaro theorem, you will see that it is necessary for the increments $$b_{n+1} - b_n$$ to be (eventually) positive. For a simple counterexample consider the following:

Let

$$a_n =\frac{1}{n}, \ \ \ \ \ \ b_n = \begin{cases} \frac{1}{n}, & \text{n even} \\\\ 0, & \text{n odd} \end{cases}$$

Then clearly $$a_n \to 0$$ and $$b_n \to 0$$. Moreover, $$\lvert b_{n+1} - b_n \rvert \geq \frac{1}{n+1}$$ and $$\lvert a_{n+1} - a_n \rvert = \frac{1}{n(n+1)}$$ implying that $$\left\lvert \frac{a_{n+1} - a_n}{b_{n+1} - b_n} \right\rvert \leq \frac{ \frac{1}{n(n+1)} } { \frac{1}{n+1} } = \frac{1}{n} \to 0.$$

However, $$\frac{a_n}{b_n} \not\to 0$$ as $$b_{2n+1} = 0$$ for all $$n \in \mathbb{N}$$. For a counterexample without $$0$$ in the denominator you just have to add a small value to the $$0$$s of $$b$$, I just wanted to keep it simple and demonstrate the idea.

As a side note: The same issues occur when using L'Hôpital's rule (which is a kind of continuous version of Stolz-Cesaro). There you have the assumption that the derivative of the function in the denominator does not vanish, which is equivalent to the function being monotone.

• In order to exclude those sorts of situations akin the counterexample you are presenting, I had explicitly restricted to $b_n$ strictly positive. However, I get your point about "For a counterexample without $0$ in the denominator ...". If I take for said small number (which also needs to $\rightarrow 0$ as $n\rightarrow \infty$) for example $1/n^{1+a}$, $a>0$, then the Stolz-Cesaro limit still exists $=0$, but the $a_n/b_n$ limit would not. Thus, answer accepted.
– Luca
Feb 14 at 19:08