I have to give a presentation on the theorem in Real Analysis with a fellow student. While I've looked over the proof and verified that, yes, step B does indeed follow logically from step A, etc. and have internalized the proof to the extent that I can replicate it myself on paper, I still feel that I have made little progress as to why this theorem works the way it does, i.e. what is the intuition behind the theorem, such that it should make sense that it follows the way it does. Therefore I ask: what kind of intuition is there about this theorem? It would also be helpful to understand how the theorem can be used effectively in analysis.
I have seen it referred to as a sort of L'Hopital's rule for sequences, which certainly seems to make sense, but I similarly have little intuitive understanding of how that rule works, either.
I am asking this question not only for my personal understanding, but also for the sake of being able to present it in an illuminating manner, such that the rest of the class can also come away with the same sort of intuition of how the theorem works and how it is useful. Any help would be greatly appreciated.
EDIT: I should point out that the formulation of the theorem we are being tasked to prove is the following:
Let $\lbrace a_n \rbrace$, $\lbrace b_n \rbrace$ be sequences, ${b_n}$ strictly increasing and unbounded. Then if $\lim\limits_{n \to \infty} \frac {a_{n+1} – a_n}{b_{n+1} – b_n} = l$ for some $l \in \mathbb R$, then $\lim\limits_{n \to \infty} \frac{a_n}{b_n} = l$ also.
Best Answer
As mentioned in the comments, one way to view it is as a discrete version of L'Hopital's rule. In analogy with a derivative of a function defined in the real numbers,
$$ \frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, $$
a "discrete derivative" of a sequence $a_n$ should be something like
$$ \frac{a_{n+h} - a_n}{h} $$
for some small $h$. Well the smallest $h$ can be in this case is $1$, so this discrete derivative should be
$$ \Delta a_n \stackrel{\text{def}}{=} \frac{a_{n+1} - a_n}{1} = a_{n+1} - a_n. $$
Now, the usual L'Hopital's rule says that if
$$ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \ell $$
then
$$ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \ell, $$
and, by analogy, the Stolz-Cesaro theorem says that if
$$ \lim_{n \to \infty} \frac{\Delta a_n}{\Delta b_n} = \ell $$
then
$$ \lim_{n \to \infty} \frac{a_n}{b_n} = \ell. $$
I also like to view the Stolz-Cesaro theorem in terms of summation. Let
$$ a_n = \sum_{k=0}^{n} \alpha_k \qquad \text{and} \qquad b_n = \sum_{k=0}^{n} \beta_k, $$
where $\beta_k \geq 0$ for all $k$ and
$$ \sum_{k=0}^{\infty} \beta_k = \infty. $$
Then the theorem says:
For illustrative purposes we will assume from here on that $\ell > 0$.
Let's introduce some new notation that should help illuminate the idea of the theorem. We'll write $f_n \sim g_n$ to mean that
$$ \lim_{n \to \infty} \frac{f_n}{g_n} = 1. $$
Intuitively, this notation says that $f_n$ and $g_n$ grow (or shrink) at the same rate. So, in terms of this new notation, the hypothesis $(1)$ becomes
$$ \alpha_n \sim \ell \beta_n. $$
That is to say, $\alpha_n$ grows or shrinks at the same rate as a constant times $\beta_n$.
The conclusion, $(2)$, becomes
$$ \sum_{k=0}^{n} \alpha_k \sim \ell \sum_{k=0}^{n} \beta_k, $$
or, pulling the constant $\ell$ inside the second sum,
$$ \sum_{k=0}^{n} \alpha_k \sim \sum_{k=0}^{n} \ell\beta_k. $$
We can just redefine $\beta_n' = \ell\beta_n$, so what the theorem is says is really:
Intuitively, if the summands $\alpha_k$ and $\beta_k$ grow at the same rate, then the sums $\sum_{k=0}^{n} \alpha_k$ and $\sum_{k=0}^{n} \beta_k$ also grow at the same rate. In other words, the sum we get when we replace $\alpha_k$ in the summand by some other equivalent sequence $\beta_k$ will behave approximately the same. We know in fact that $\sum_{k=0}^{\infty} \beta_k$ diverges, so we can also interpret this as saying that the partial sums diverge at the same rate.
Appendix: If $\ell = 0$ then $\alpha_n/\beta_n \to \ell = 0$ can be interpreted to mean that $\alpha_n$ is "smaller" than $\beta_n$ in the limit, and the conclusion that $$ \sum_{k=0}^{n}\alpha_k \left/ \sum_{k=0}^{n}\beta_k \right. \to \ell = 0 $$ can be interpreted to mean that $\sum_{k=0}^{n}\alpha_k$ is therefore "smaller" than $\sum_{k=0}^{n}\beta_k$ in the limit.
Perhaps this (contrived) example will give some idea of the usefulness of this theorem. Suppose we wish to calculate
$$ \lim_{n \to \infty} \log(n)^{-1} \sum_{k=1}^{n} \sin(1/k). $$
We could begin by noticing that $\sin(1/k) \sim 1/k$ as $k \to \infty$, so by Stolz-Cesaro we know that
$$ \sum_{k=1}^{n} \sin(1/k) \sim \sum_{k=1}^{n} \frac{1}{k}. $$
This second sum is much easier to estimate. From this we know that
$$ \log(n+1) = \int_1^{n+1} \frac{dx}{x} \leq \sum_{k=1}^{n} \frac{1}{k} \leq 1 + \int_1^n \frac{dx}{x} = 1 + \log(n), $$
so
$$ \sum_{k=1}^{n} \frac{1}{k} \sim \log(n). $$
Consequently,
$$ \sum_{k=1}^{n} \sin(1/k) \sim \log(n). $$
That is,
$$ \lim_{n \to \infty} \log(n)^{-1} \sum_{k=1}^{n} \sin(1/k) = 1. $$
Because of Stolz-Cesaro, we know that instead of having to deal with the troublesome summand $\sin(1/k)$ we could replace it with the much simpler summand $1/k$ and not change the limiting behavior of the sum.