I would like to know how to evaluate $$\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$
There are a couple of issues I have with this. Firstly, depending on the value of $x$, it seems, at least numerically, that this value is always
$$\pm\sqrt{\frac{\sqrt5-1}{2}}$$
If this is true, how would I prove this?
Secondly, I tried using complex numbers to evalaute this, as shown briefly below:
$$C=\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\cos(2r-1)x}{2r-1}$$
$$S=\sum\limits_{r=1}^\infty(-1)^{r+1}\frac{\sin(2r-1)x}{2r-1}$$
$$\implies C+iS=\arctan e^{ix}$$
on using the power series of $\arctan x$. Differentiating yields a completely imaginary number:
$$\frac{i}{2}\sec x$$
which proves that the value of $C$ is a constant. However, this gives me no information whatsoever on the value of $C$.
Thank you for your help.
Best Answer
As you noted, the real part of the sum, $C$, is constant no matter what the value of $x$ (so long as $x$ is real). Thus plug in $x = 0$
$$C = \operatorname{Re}(\arctan e^{i0}) = \frac{\pi}{4}$$