Does the product of moment generating functions of IID variables approach that of a normal distribution? When you sum N IID samples from a univariate distribution, the sum approaches a normally-distributed value.
Likewise, when you sum variables together, the moment-generating functions of their distribution can be multiplied to produce the resulting moment-generating function of the sum.
It would seem that you could exponentiate a variable's moment-generating function by a large N, and the resulting function would be close to that of a normal distribution's.
However, from a mathematical perspective, this makes no sense, as you are simply 
For simplicity's sake, let's use variables from continuous distributions that have an MGF (to avoid problems with the domain).
One example I am having trouble with is the Chi-squared distribution, with an MGF of 
(1-2t)^(k/2)

The resulting MGF of the sum of many of these variables would be 
(1-2t)^(N*k/2)

If we assume N*k/2 is an integer, C, then the MGF is
(1-2t)^C = 1-2t * C + 4t^2 * C*(C-1)/2 +...

Because C ≈ C-1 ≈ C-2 ≈ ... for large C, this can simplify to 
sum((-2t*C)^n/n!) from n=0 to C ≈ e^(-2t*C)

Of course, the later terms deviate more strongly from the normal distribution's, but those derivatives don't affect the function's value very much.
Unfortunately, this MGF is degenerate, because it simply describes that of a variable that is always C. There is no t^2 term in the exponent, which is required for a normal MGF. Did I make a mistake in my calculation, or is there some fundamental assumption that I overlooked/violated?
Edit: I replaced "added" with "multiplied" in one instance because I was referring to multiplying the MGFs, not adding them.
Edit 2: I realized I forgot the minus sign in the exponent of the Chi-squared MGF. That resulted in a computational error.
 A: 
When you sum $N$ IID samples from a univariate distribution, the sum approaches a normally-distributed value.

Not exactly. The sum doesn't converge. Under certain conditions, the average converges to the mean, and a scaled version of the de-meaned average converges to a normal distribution.
In particular, using your example, say you have an RV $X_i \sim \chi^2$, then its MGF is
$$
\left( 1 - 2t \right) ^{-\frac{k}{2}}. \; (1)
$$
The second-order expansion of (1) is 
$$
\sim \left( 1 - 2t \right) ^{-\frac{k}{2}}|_{t = 0} + 
\frac{\partial \left( 1 - 2t \right) ^{-\frac{k}{2}}}{\partial t}|_{t = 0}t + 
\frac{\partial^2 \left( 1 - 2t \right) ^{-\frac{k}{2}}}{2 \partial t^2}|_{t = 0}t^2
=
1 + k t + \frac{k}{2}(k + 2) t^2
$$
If you have $n$ independent RVs $X_1, \ldots, X_n$ with this distribution, then the MGF of $\sum_{i = 1}^n\left[x_i/n\right]$, which is just a linear combination, is therefore known to be
$$
\sim \left(  1 + k\frac{t}{n} + \frac{k}{2}(k + 2)\left(\frac{t}{n}\right)^2 \right)^n
$$
The MGF of $\sqrt{n} \left(\sum_{i = 1}^n\left[x_i/n\right] - k\right)$, therefore, 
is
$$
\sim \left(  1 + \frac{\frac{k}{2}(k + 2)t^2}{n} \right)^n \sim e^{\frac{k}{2}(k + 2)t^2}
$$
This is the MGF of a normal distribution.
