I am trying to go over a proof of the CLT given by this site.
I understood everything up to the point where the expansion for the logarithm was used:
$$x=\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3).$$ This was used to expand: $$n\ln\left(1+\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)\ +\cdots\right).$$
How is this correct if the above is not in the form of $\ln(1+x)$? Am I missing something?
The logarithm of the moment generating function is
$$\ln m_u(t)=n\cdot \ln\left(1+\underbrace{\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)}_{\color{red}x} +\ldots\right)$$
Now you set $x$ equal to $\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3) \quad (*)$
Thus you can use
$$\ln m_u(t)=n\cdot \ln\left(1+x\right)$$
Using the series expansion of $\ln(x+1)$
$$n\cdot \ln m_u(t)=n\cdot \ln\left(1+x\right)=n\cdot \left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots\right)$$
Now you insert the expression for $x$, see (*). I do it for the first three summands only.
$n\cdot \ln m_u(t)$
$$=\color{blue}n\cdot \left(\frac{t^2}{2\color{blue}n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)-\frac{\left(\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)\right)^2}{2}+\frac{\left(\frac{t^2}{2n}\ +\ \frac{t^3}{3! n^{\frac{3}{2}}}E(U_i^3)\right)^3}{3}+\ldots\right)$$
You see at the first summand $n$ is cancelling out only.
In the other cases the exponents of $n$ in the denominators are larger than 1 $\left(\frac{3}{2}, 2, 1+\frac{3}{2} , 3, ... \right)$.
Consequently for $n\to \infty$ the summands become $0$, exept the first one.