I'm trying to prove some random variable converges to a constant $c$ and this limit showed up
$$\lim_{n \to \infty}\left(1+c\cdot \left(\exp\left(\frac{it}{n}\right) – 1\right)\right)^n$$
For the random variable in question to converge to $c$ it's necessary that this limit above be equal to $\exp(itc)$, which is the reason for the title. I tried using the classic:
$$\lim_{n \to \infty}\left(1+\frac{z}{n}\right)^n=e^z$$
but this did not help me. I'm thinking I'm gonna have to use some other known limit involving the exponential but I don't know how to do so. I appreciate any help!
Best Answer
Here is a sketch of a proof of the main ingredient: $$\big(1+\frac{c_n}{n}\big)^n\xrightarrow{n\rightarrow\infty} e^c$$ for any $c_n\in\mathbb{C}$ with $c_n\xrightarrow{n\rightarrow\infty}c$.
Following Rick Durrett's approach to the classical CLT
Lemma I: Let $\{z_j,w_j:1\leq j\leq n\}\subset\mathbb{C}$ such that $\max_j\{|z_j|,|w_j|\}\leq \theta$. Then \begin{align} \Big|\prod^n_{j=1}z_j-\prod^n_{j=1}w_j\Big|\leq \theta^{n-1}\sum^n_{j=1}|z_j-w_j| \end{align}
Proof: Let $a_0=\prod^n_{j=1}w_j$ and for $k\geq1$, $a_k=\Big(\prod^k_{j=1}z_j\Big)\Big(\prod^n_{j=k+1}w_j\Big)$. Then \begin{align} \Big|\prod^n_{j=1}z_j-\prod^n_{j=1}w_j\Big|&=\Big|\sum^n_{j=1}(a_j-a_{j-1})\Big|\leq\sum^n_{j=1}|a_j-a_{j-1}|\\ &=\sum^n_{j=1}|z_1\cdot\ldots\cdot z_{j-1}(z_j-w_j)w_{j+1}\cdot\ldots\cdot w_n|\leq \theta^{n-1}\sum^n_{j=1}|z_j-w_j|. \end{align}
Lemma II: Suppose $|z|\leq1$, then $|e^z-1-z|\leq|z|^2$.
Proof Observe that $2^{n-1}\leq n!$ for $n\geq2$. Since $e^z-1-z=\sum_{n\geq2}\frac{z^n}{n!}$ and $|z|\leq 1$, it follows that \begin{align} |e^z-1-z|\leq \frac{|z|^2}{2}\sum_{n\geq2}2^{-(n-2)}=|z|^2 \end{align}
finally: Proposition: Suppose that $\{c_n\}\subset\mathbb{C}$ and that $c_n\rightarrow c$. Then, $\big(1+\frac{c_n}{n}\big)^n\rightarrow e^c$.
Proof Let $\gamma>|c|$ and $n_0$ large enough so that $|c_n|<\gamma$ and $\gamma/n\leq 1$ whenever $n\geq n_0$. If $z_j=1+\frac{c_n}{n}$ and $w_j=e^{c_n/n}$, $1\leq j\leq n$, then have that \begin{align} \max_{1\leq j\leq n}\{|z_j|,|w_j|\}&\leq& e^{\gamma/n}\\ \big|e^{c_n/n}-1-\frac{c_n}{n}\big|&\leq& \frac{\gamma^2}{n^2}. \end{align} Therefore \begin{align} \big|(1+\frac{c_n}{n})^n-e^{c_n}\big|\leq e^{\gamma (n-1)/n}n\frac{\gamma^2}{n^2}\rightarrow0. \end{align} The conclusion follows from the continuity of the exponential function.