I'm trying to proof the Central Limit Theorem. Unless you have a better suggestion, what I'm doing is show that the moment function of the normal distribution, which is exp(t^2/2) is the same that the moment function for a distribution f(x), with known mean u and standard deviation s. This procedure is sketched (but not developed) in the book "Probabilities and Statistics", 4th Ed. by Ronald Walpole and Raymond Myers. (Chapter 6, section 8)

However I got some problems in this last procedure: The way to make the proof is with a McLaurin series of the ln of the moment. And the moment is also represented with a series.

The series for the moment is:
Mx(t/s*sqrt(n)) = 1 + u't/s*sqrt(n) + u''t^2 /2! s^2 * n +....

And the series of ln is well known:

ln(1+ v) = v - v^2/2 + v^3/3 -...

So I take only the first 2 terms of each series; I replace the first 2 terms of Mx in ln(1+v) and get:

u*t/s*sqrt(n) - (t/s*sqrt(n))^2/2

The first term is nulled by another term that appears somewhere in the process. But the problem is the second term: is negative, and must be positive. I triple-checked the process and there is not any sign error. The term is negative because the ln series is negative in the even terms. Also another problem is that s never goes away, because it's present since I input it in Mx(t/s*sqrt(n)). (n does disappear, since the series is multiplied by n and the second term is squared, thus I get n/sqrt(n)^2 = 1)

If I got a way to make the term positive and also make disappear s, I would get t2/2, which passed to an exp() function results into the same moment for the normal: exp(t^2/2)

Sorry for not providing the full procedure, it's full of signs and it's hard to copy it in this limited forum.

Any suggestions?