I asked this question on MSE here but I was told it would do better here

I always wonder how mathematicians proved theorems before Cauchy’s epsilon-delta proof. Since many "recent" theorems, such as the Stolz-Cesàro theorem, Cauchy limit theorem , Cauchy D'Alembert's Law, use epsilon-delta as part of their proofs, I wonder how Euler would have proved Stolz-Cesàro theorem for example. Although this theorem was not known to Euler, I am pretty sure that he could have "proved" it in minutes.

Stolz-Cesàro theorem case $\frac{*}{\infty}$:- If $b_n $ is a monotone increasing sequence and $\lim \limits_{n \to \infty} b_n = \infty $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

Stolz-Cesàro theorem case $\frac{0}{0}$:- If $b_n $ is a monotone decreasing sequence and $\lim \limits_{n \to \infty} b_n = \lim \limits_{n \to \infty} a_n = 0 $, and if $\lim \limits_{n \to \infty} \frac{a_{n+1}-a_n}{b_{n+1}- b_n}= l \in \overline{\mathbb{R}} $, then $\lim \limits_{n \to \infty} \frac{a_n }{b_n}=l$.

Cauchy D'Alembert's Law If $a_n$ is a sequence of real numbers and if $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n} =l$ the $\lim\limits_{n \to \infty} \sqrt[n]{a_n}=l$

I want to see how to "prove" one of these theorems without rigor.


1 Answer 1


The "standard" way of proving such things without epsilontics would be to use infinitesimal analysis, at which Euler was an expert of course. From this point of view, the formula $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\ell$ means that for all unlimited (informally, infinite) values of $n$, the ratio $\frac{a_{n+a}}{a_n}$ is infinitely close to $\ell$, and in fact it is enough to require such a relation of infinite proximity $\frac{a_{n+a}}{a_n}\approx\ell$ only for $n$ greater than some fixed unlimited $H>0$.

Since you are interested in arguments "without rigor", I will present a plausible argument that requires further work if one wants to make it fully OK. Suppose $H=K^2$ where $K$ is of course still unlimited. Then $K$ is "much" smaller than $H$. Since the value on the limit is only dependent on the behavior of an infinite tail of the sequence, we can modify the original sequence by replacing the first $K$ terms in such a way that all the ratios $\frac{a_n}{a_n}$ are precisely $\ell$, without appreciably affecting either the limit of the ratios or the limit of the root. What we obtain is that the sequence is appreciably of the form $C\ell^n$ (again, up to modifications of some initial terms). Then the $n$-th root for $n>H$ will be appreciably $\sqrt[n]{C} \, \sqrt[n]{\ell^n}\approx \ell$.

  • 1
    $\begingroup$ A very good example! :) $\endgroup$ Commented Jan 1 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.