Showing that $-\frac{1}{2}=\sum^{\infty}_{n=1}\frac{(-1)^{n}\sin(n)}{n}$ So I've been trying to prove that
$$\sum^{\infty}_{n=1}\frac{(-1)^{n}\sin(n)}{n}=-\frac{1}{2}$$
I've tried putting various bounds on it to see if I can "squeeze" out the result. Say something like (one of many tried examples):
$$ -\frac{1}{n}-\frac{1}{2}\leq \sum^{n}_{k=1}\frac{(-1)^{k}\sin(k)}{k}\leq \frac{1}{n}-\frac{1}{2}$$
I've tried too see if I could find some periodic continuous function in order to use Parseval's theorem, but I can't come up with any that work.
May I please get a hint or some piece of the puzzle for this problem?
 A: Guide:
Verify by computing the fourier series of $f(x) = \frac{x}{\pi}, -\pi < x<\pi$
$$\frac{x}{\pi} = \frac{2}{\pi}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\sin(nx)$$
and choose an appropriate value of $x$.
A: A direct approach:
$$\sum_{n=1}^\infty\frac{(-1)^n\sin n}n=\sum_{n=1}^\infty\frac{(-1)^n\Im (e^{in})}n=-\Im\sum_{n=1}^\infty\frac{(-1)^{n+1}(e^i)^n}n=-\Im\ln(1+e^i)$$
where the Maclaurin series of $\ln(1+z)$ was used. As can be seen by drawing a diagram, $\arg(1+e^i)=\frac12$. This is thus the imaginary part of the logarithm, and thus the original sum is $-\frac12$.
A: Its is also possible to calculate the series above using the residue theorem from complex analysis. It goes as follows:$$\sum_{i=-\infty}^\infty \text{Res}\left(\pi\csc(\pi z)\cdot\frac {\sin(z)}z,n\right)=-\text{Res}\left(\pi\csc(\pi z)\cdot\frac {\sin(z)}z,0\right)$$
The LHS becomes: 
$$\sum_{i=-\infty}^\infty\lim_{z\to n}\left[\pi  (z-n)\csc(\pi z)\cdot\frac {\sin(z)}z\right]=\sum_{i=-\infty}^\infty(-1)^n \frac {\sin(n)}n=2\sum_{i=1}^\infty(-1)^n \frac {\sin(n)}n$$
The RHS becomes: $$\lim_{z\to 0}\left[\pi z\csc(\pi z)\cdot\frac {\sin(z)}z\right]=1$$
Putting all together (LHS $=-$ RHS) you get: $$\sum_{i=1}^\infty(-1)^n \frac {\sin(n)}n=-\frac 12$$
