Asymptotics of a sequence of exponentials How can one show that 
\begin{equation}
\big|1 - e^{i(\frac{1}{n+1} - \frac{1}{n})}\big| = \frac{1}{n^2} + o\big(\frac{1}{n^2}\big)
\end{equation}
for all $n \in \mathbb{N}$?
I tried using Taylor series, but there is a factor with an $i$, I then tried writing out the modulus but I can't seem to obtain this result. 
 A: Well, first of all we can write:
$$\left|1-\exp\left(\theta\cdot i\right)\right|=\sqrt{\left(1-\cos\left(\theta\right)\right)^2+\left(-\sin^2\left(\theta\right)\right)}=2\cdot\left|\sin\left(\frac{\theta}{2}\right)\right|\tag1$$
When $\theta\in\mathbb{R}$
A: the derivative of $\exp$ at $0$ is $1$, so $\lim_{y \to 0} \frac {\exp(y)-1}{y} = 1$ .
Plug in $y_n = i(\frac 1{n+1} - \frac 1n) = \frac 1{in(n+1)}$ to get $\lim_{n \to \infty} in(n+1)(\exp(y_n)-1) = 1$.
Since $\lim_{n \to \infty} \frac n{n+1} = 1$, you can multiply them to get $\lim_{n \to \infty} in^2(\exp(y_n)-1) = 1$.
Then taking modulus (which is continuous), you get $\lim_{n \to \infty} n^2|\exp(y_n)-1| = 1$.
Now you are done because this means $\lim_{n \to \infty}n^2(|\exp(y_n)-1|-\frac 1{n^2}) = 0$, so $|\exp(y_n)-1|-\frac 1{n^2} = o(\frac1 {n^2})$
A: Let
$$\theta=\frac{1}{n+1} -\frac1n=-\frac{1}{n(n+1)}$$
since
$$e^{i\theta}=1+i\theta+io(\theta)$$
$$\big|1 - e^{i\theta}\big|=\big|1 - 1-i\theta+io(\theta)\big|=|\theta| + o(|\theta|)$$
thus
$$\big|1 - e^{i(\frac{1}{n+1} - \frac{1}{n})}\big| =\frac{1}{n(n+1)}+o\left(\frac{1}{n(n+1)}\right)=\frac{1}{n(n+1)}+o\left(\frac{1}{n^2}\right)$$
