Where is Cauchy's wrong proof? Allegedly, Cauchy mistakingly "proved" that pointwise convergence of continuous functions is continuous. I saw this somewhere in a book, and it is also in wikipedia: 

Uniform convergence. In his Cours d'analyse of 1821, Cauchy "proved" that if a sum of continuous functions converges pointwise, then its limit is also continuous. However, Abel observed three years later that this is not the case. For the conclusion to hold, "pointwise convergence" must be replaced with "uniform convergence".1 There are many counterexamples. For example, a Fourier series of sine and cosine functions, all continuous, may converge to a discontinuous function such as a step function.

Well, I searched in the book mentioned (or rather, this one ) and didn't find it. Where is it?
 A: In the première partie, chapitre VI, $\S$1, the 1er Théorème is:

Lorsque les differens [sic] termes de la série (1) sont des fonctions d'une même variable $x$, continues par rapport à cette variable dans le voisinage d'une valeur particulière pour laquelle la série est convergente, la somme $s$ de la série est aussi, dans le voisinage de cette valeur particulière, fonction continue de $x$.

(1) is simply a sequence $u_0,u_1,\ldots$.  
This says that an infinite sum of functions each continuous in a neighbourhood of point is also continuous in a neighbourhood of that point.
Cauchy's argument, slightly restated, is this: let the $u_i$ be functions of a variable $x$ continuous on some neighbourhood of a point $X$. Let $s_n$ be their $n$th partial sum, $s$ the sum and $r_n=s-s_n$. He notes that $s_n$ is continuous in a neighbourhood of $X$.  Let $\alpha$ be infinitesimal and consider the effect on $s(X),s_n(X),r_n(X)$ when we increase $X$ by $\alpha$. For each $n$, since $s_n$ is continuous, $s_n(X+\alpha)-s_n(X)$ is infinitesimal.  Since $s_n \to s$ it follows that $r_n$ is infinitesimal for large $n$.  He then says that the difference $r_n(X+\alpha)-r_n(X)$ "will become imperceptible" (deviendra insensible -- this isn't how he usually refers to infinitesimal quantities) at the same time (i.e. for the same values of $n$) that $r_n$ itself does.  Since for every $n$ we have $s=s_n+r_n$, it follows by letting $n$ take a value trés-considérable that the increase of $s$ is infinitesimal and so $s$ is continuous.
The error is quite subtle and gives a good idea as to why modern formalisms of NSA were needed to untangle which infinitesimal arguments are valid.  We have two notions of continuity: the ordinary $\epsilon\delta$ one, and what I'll call NS-continuity: $f$ is NS-continuous at the point $x$ if for all $y$ infinitely close to $x$ we have $f(x)-f(y)$ infinitesimal.  These are not always equivalent (example: if $N$ is an infinitely large positive integer then $f(x)=x^N$ is continuous but not NS-continuous at 1).  While the $s_n$ are continuous for all, even infinitely large, $n$ they may not be NS-continuous, and this is where the argument fails.
