We can't we examine the pointwise limit of the characteristic functions for the sample means to prove the central limit theorem? My Latex skills are horrible, and I'm unable to type my problem out in its full glory on this interface, but I believe I'll be able to sufficiently communicate my concerns to you all.
A problem in a textbook that I'm reading asked the reader to prove the central limit theorem for sample means using characteristic functions.
The characteristic function I got for the sample mean is $(E(e^{itX/n}))^n$ where $X$ is a random variable with mean $\mu$ and variance $\sigma^2$.
Once you expand $e^{itX/n}$ using Taylor series, apply expectation, raise the result to the $n^{th}$ power, and then take limits, the limit I get is $e^{it\mu}$ which is clearly incorrect.
Can someone explain to my why this doesn't work?
Thank you!
 A: Let $\mu_{n} = \frac{S_{n} - E[S_{n}]}{\sqrt{D[S_{n}]}}$. Now, to prove the central limit theorem, we have to show that the characteristic function of $\mu_{n} \rightarrow e^{-t^2 / 2}$. Let $a = E[\xi_{i}], \; \sigma^{2} = D[\xi_{k}]$. Now, let's transform $\mu_{n}$:
$$
\mu_{n} = \frac{\sum_{i = 1}^{n} \xi_{k} - E[\xi_{k}]}{\sqrt{\sum_{i = 1}^{k} D[\xi_{k}]}} = \sum_{i = 1}^{n} \frac{\xi_{k} - a}{\sqrt{n} \cdot \sigma} = \sum_{i = 1}^{n} \frac{\eta_{k}}{\sqrt{n}}
$$
Here $\eta_{k} = \frac{\xi_{k} - a}{\sigma}$.
Now, note that $E[\eta_{k}] = 0, D[\eta_{k}] = 1 \rightarrow E[\eta_{k}^2] = 1$
So, the characteristic function of $\eta_{k}$ is $\phi_{\eta_{k}}(t) = 1 - \frac{t^2}{2} + o(t^{2})$
So,
$$\phi_{\mu_{n}}(t) = E \left[\exp \left(i \cdot t \cdot \sum_{i = 1}^{n} \frac{\eta_{k}}{\sqrt{n}}\right) \right ] = (because \; \eta_{k} \; are \;  independent)  = \prod\limits_{k = 1}^{n} \phi_{\eta_{k}} \left(\frac{t}{\sqrt{n}} \right) = \left(1 - \frac{t^2}{2n} + o(t^2 / n) \right)^{n} \rightarrow e^{-\frac{t^2}{2}}$$
So $\mu_{n} \xrightarrow[]{d} N (0, 1)$
