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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.

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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 $\sqrt[n]{C} \, \sqrt[n]{\ell^n}\approx \ell$.

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

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