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?

Update following Stefan's Answer counterexample:

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).


1 Answer 1


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:


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

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.

  • $\begingroup$ 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. $\endgroup$
    – Luca
    Feb 14 at 19:08

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