Proving this partial sum is bounded I want to show that the partial sum of $\sum_{n=1}^{\infty} {{e^{inx}}}$ with $x ∈ {(0, 2{\pi})}$ is bounded.
I know that the solution is: $|\frac {1-e^{i(n+1)x}}{1-e^{ix}} - 1| \leq \frac {2}{|1-e^{ix}|} +1 $. Thus, it is indeed bounded.
But I don't understand how you achieve this solution step by step and I really want to understand. I know that you probably have to use the formular: $\sum_{n=1}^{N} {{x^{n}}} = \sum_{n=0}^{N-1} {{x^{(n+1)}}} = |\frac {1-x^{(n+1)}}{1-x}|$. But why is it $- 1$ in the solution and then $\leq |\frac {2}{1-e^{ix}}| +1$? 
I would be very thankful if someone could explain me the steps to reach this solution.
 A: It's the formula for the sum of a GP: $\sum_{n=1}^N z^n=z(z^N-1)/(z-1)$.
Here $z=e^{ix}$
and so
$$\sum_{n=1}^N e^{inx}=e^{ix}\frac{e^{iNx}-1}{e^{ix}-1}.$$
We can re-arrange this to bring in the sine function:
$$\sum_{n=1}^N e^{inx}=e^{i(N+1)x/2}\frac{e^{iNx/2}-{e^{-iNx/2}}}{e^{ix/2}-e^{-ix/2}}=e^{i(N+1)x/2}\frac{\sin(Nx/2)}{\sin(x/2)}.$$
As $|e^{i(N+1)x/2}|=1$ and $|\sin(Nx/2)|\le1$ then
$$\left|\sum_{n=1}^N e^{inx}\right|\le\frac{1}{\sin(x/2)}.$$
(If $0<x<2\pi$ then $\sin(x/2)>0$.)
A: We have $|e^{i\alpha}|=|\cos\alpha+i\sin\alpha|=1$ and it's clear 
$$\left|\frac {1-e^{i(n+1)x}}{1-e^{ix}} - 1\right| = \left|\frac {e^{ix}-e^{i(n+1)x}}{1-e^{ix}} \right|= \left|e^{ix}\frac {1-e^{inx}}{1-e^{ix}} \right| \leq |e^{ix}|\frac {1+|e^{inx}|}{|1-e^{ix}|}  \leq \frac {2}{|1-e^{ix}|} +1$$
but the number $1$ appears in formula isn't a rule to someone use it everytime want. Indeed we know that with $1$ we can simplify our formule because it appeared in both numerator and denominator.
A: Okay, I came up with the solution to this problem.
Since $e^{ix}$ is a constant, we can rewrite $\sum_{n=0}^N e^{inx}=e^{{ix}^{n}}$. 
Thus the formular gives us: $\sum_{n=0}^N e^{{ix}^{n}}=|\frac{e^{{ix}^{N+1}}-1}{e^{ix}-1}|$, where ${e^{ix}-1} \neq 0$. 
The $-1$ in the solution is used because the index starts at $1$ instead of $0$. 
Now we use $|\frac{a}{b}| = \frac{|a|}{|b|}$. Since the denominator is a constant we focus now on the numerator.
$|{e^{{ix}^{N+1}}-1}| \leq |({e^{ix})^{N+1}}| + |-1|.$
We know that $e^{ix}$ is part of the unit circle, thus $|e^{ix}| = 1.$
This means   $|{e^{{ix}^{N+1}}}| = |{e^{ix}|^{N+1}} = 1$.
Thus, $|({e^{ix})^{N+1}}-1| \leq 2$. Which results in $\leq \frac {2}{|1-e^{ix}|}$. Now in the task we just have to keep in mind that the index starts at $1$. We have now shown that the sum is indeed bounded.
