I'm struggling with this problem,
Find an expression for $\sin{x} + 2\sin(2x) +3\sin(3x) + ... +n\sin(nx)$
The problem states that I have to explicitly show that this series can be expressed as
$$\frac{(n+1)\sin(nx) - n\sin((n+1)x)}{\sin^2(x/2)}$$
For all $ n\in\mathbb{N}$ and all $x\in\mathbb{R} $
Hint:
This is the imaginary part of $$\mathrm e^{ix}+2\,\mathrm e^{2ix}+\dots+n\,\mathrm e^{nix},$$ which is ‘almost’ the derivative of $$1+\mathrm e^{ix}+\mathrm e^{2ix}+\dots+\mathrm e^{nix}.$$ The last expression is a geometric series.
Variant: you may already know these formulæ (which can be obtained in a way similar to the first method): $$\sum_{k=1}^n \sin kx=\frac{\sin\dfrac{(n+1)x}{2}}{\sin\dfrac x2}\,\sin\dfrac{nx}{2},\qquad\sum_{k=1}^n \cos kx=\frac{\sin\dfrac{(n+1)x}{2}}{\sin\dfrac x2}\,\cos\dfrac{nx}{2} $$ and differentiate the second of these.
Hint:
With $z=e^{ix}$, your sum is the imaginary part of
$$\sum_{k=1}^n kz^k=z\frac{1 - z^n}{(1-z)^2} - nz\frac{z^n}{1-z}=\frac{2z}{(1-z)^2} +\frac{nz}{1-z}.$$
The rest is routine work.
Hint: $$ \sum_{k=1}^n k\sin\left(kx\right)=\Im\left(\sum_{k=1}^n ke^{kix}\right) $$ using $e^{ix}=\cos(x)+i\sin(x)$. Now, we can rewrite the sum as a double sum where $$ \Im\left(\sum_{k=1}^n ke^{kix}\right)=\Im\left(\sum_{m=1}^n\sum_{j=m}^n e^{jix}\right)=\Im\left(\sum_{m=1}^ne^{mix}\sum_{j=0}^{n-m} e^{jix}\right). $$ The inner sum is a geometric sum, so this simplifies to $$ \Im\left(\sum_{m=1}^ne^{mix}\sum_{j=0}^{n-m} e^{jix}\right)=\Im\left(\sum_{m=1}^ne^{mix}\left(\frac{1-e^{(n-m+1)ix}}{1-e^{ix}}\right)\right). $$ This is, again, geometric (actually two geometric sums), and you can continue the computation.
