Question about the proof of Central Limit Theorem My instructor proved the central limit theorem using the characteristic function. I think the proof is a standard one because I found basically the same proof in wikipedia.
So for i.i.d. ${X_1, X_2,\cdots, X_n}$ with $E[X_i]=\mu$ and $\text{Var}(X_i)=\sigma^2<+\infty$, he then defined
$$Y_i=\frac{X_i-\mu}{\sqrt{n}}$$
Then $E[Y_i]=0$ and $\text{Var}(Y_i)=\sigma^2/n$.
So the characteristic function of $Y_i$ is
$$\phi(k)=1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)$$
Then he further define
$$Z=Y_1+Y_2+\cdots+Y_n$$
and so the characteristic function of $Z$ is
$$\Phi(k)=\phi(k)^n=\left(1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)\right)^n$$
Then he claimed that when $n\rightarrow\infty$
$$\Phi(k)=\lim_{n\rightarrow\infty}\left(1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)\right)^n=\exp\left(-\frac{k^2\sigma^2}{2}\right)$$
Performing the inverse Fourier transform we get
$$P(Z)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{Z^2}{2\sigma^2}\right)$$
Then finally defining
$$\tilde{X}=\frac{Z}{\sqrt{n}}+\mu=\frac{X_1+X_2+\cdots+X_n}{n}$$
it can be shown that
$$P(\tilde{X})=\frac{1}{\sqrt{2\pi}(\sigma/\sqrt{n})}\exp\left(-\frac{(\tilde{X}-\mu)^2}{2(\sigma/\sqrt{n})^2}\right)$$
I found slightly different proof in wiki but the basic idea is the same.
Now where I have problem is the step
$$\lim_{n\rightarrow\infty}\left(1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)\right)^n=\exp\left(-\frac{k^2\sigma^2}{2}\right)$$
I know that I cannot stupidly put the limit inside the bracket for otherwise I'd get the answer $1^n=1$. But why do I know I cannot do it to the $O(1/n)$ term, but can do it to the $o(1/n)$ terms?
Are there something I am missing?
 A: When we take the logarithm, we have to be careful, since the little "o" is complex (at least we have to choose a good determination). Alternatively, we can use and show the fact that if $\left(z_n\right)_{n\geqslant 1}$ is a sequence of complex numbers which converges to $z$, then 
$$\lim_{n\to +\infty}\left(1+\frac{z_n}n\right)^n=e^z.$$
To see this, notice that if $(x_n)_{n\geqslant 1}$ and $(y_n)_{n\geqslant 1}$ are two sequences of complex numbers such that $\max\left\{|x_n|,|y_n|\right\}\leqslant R$, then 
$$\tag{*}\left|\prod_{j=1}^nx_j-\prod_{j=1}^ny_j\right|\leqslant R^{n-1}\sum_{j=1}^n\left|x_j-y_j\right|$$
(this can be done by induction). 
Now, define $x_n:=\left(1+\frac{z_n}n\right)^n$ and $y_n:=\left(1+\frac{z}n\right)^n$. By (*) with $R:=1+\sup_{j\geqslant 1}|z_j|/n$, we get that 
$$\left|\left(1+\frac{z_n}n\right)^n-\left(1+\frac{z}n\right)^n\right|
\leqslant \left(1+\sup_{j\geqslant 1}|z_j|/n\right)^{n-1}\left|z_n-z\right|,$$
which gives what we wanted, as $\left(1+\sup_{j\geqslant 1}|z_j|/n\right)^{n}$ converges to $\exp\left(\sup_{j\geqslant 1}|z_j|\right)$.
A: To prove that if a complex sequence
$$\lim_{n\rightarrow+\infty}z_n=z,$$
then
$$\lim_{n\rightarrow+\infty}\left(1+\frac{z_n}{n}\right)^n=e^z$$
Proof:
Let $(x_j)_{n\ge j \ge 1}$ and $(y_j)_{n\ge j \ge 1}$ be two sequences of complex numbers.
Let $|x_j| \le R$ and $|y_j| \le R$ for all $1 \le j \le n$.
Then
$$\bigg|\prod^n_{j-1}x_j-\prod^n_{j=1}y_j\bigg|\le R^{n-1}\sum_{j=1}^n|x_j-y_j|.$$
Let's prove the above by induction.
For $n=1$, it is obviously true.
Suppose
$$\bigg|\prod^n_{j-1}x_j-\prod^n_{j=1}y_j\bigg|\le R^{n-1}\sum_{j=1}^n|x_j-y_j|$$
Then
$$\bigg|\prod^{n+1}_{j-1}x_j-\prod^{n+1}_{j=1}y_j\bigg|=\frac{1}{2}\bigg|\left(x_{n+1}-y_{n+1}\right)\left(\prod_{j=1}^nx_j+\prod_{j=1}^ny_j\right)+\left(x_{n+1}+y_{n+1}\right)\left(\prod_{j=1}^nx_j-\prod_{j=1}^ny_j\right)\bigg|$$
$$\le \frac{1}{2}|x_{n+1}-y_{n+1}|\bigg|\prod_{j=1}^nx_j+\prod_{j=1}^ny_j\bigg|+\frac{1}{2}|x_{n+1}+y_{n+1}|R^{n-1}\sum_{j=1}^n|x_j-y_j|$$
$$\le \frac{1}{2}|x_{n+1}-y_{n+1}|2R^n+\frac{1}{2}2RR^{n-1}\sum_{j=1}^n|x_j-y_j|$$
$$= R^n\sum_{j=1}^{n+1}|x_j-y_j|$$
Now for fixed $n$, let
$$x_j=1+\frac{z_n}{n}$$
$$y_j=1+\frac{z}{n}$$
$$R=1+\frac{\sup_{i\ge 1}|z_i|}{n}$$
Then we have
$$\bigg|\left(1+\frac{z_n}{n}\right)^n-\left(1+\frac{z}{n}\right)^n\bigg|\le \left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^{n-1}|z_n-z|$$
Take limit $n \rightarrow +\infty$, we have
$$\lim_{n\rightarrow+\infty}\bigg|\left(1+\frac{z_n}{n}\right)^n-\left(1+\frac{z}{n}\right)^n\bigg| \le \lim_{n\rightarrow+\infty}\left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^{n-1}|z_n-z|$$
$$=\lim_{n\rightarrow+\infty}\left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^n \left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^{-1}|z_n-z|$$
$$=\exp\left(\sup_{i\ge 1}|z_i|\right) \times 1 \times 0$$
$$=0$$
Hence
$$\lim_{n\rightarrow+\infty}\left(1+\frac{z_n}{n}\right)^n=\lim_{n\rightarrow+\infty}\left(1+\frac{z}{n}\right)^n$$
$$=e^z$$
