Evaluating $\sum_{k=0}^\infty\sin(kx)$ and $\sum_{k=0}^\infty\cos(kx)$

Question

Playing with sines I wanted to find $$ S(x)=\sum_{k=0}^\infty\sin(kx) $$

Writing it as $$ S(x)=\mathrm{Im}\big(A(x)\big),\quad\text{where}\quad A(x)=\sum_{k=0}^\infty e^{ikx} $$ and using $z=e^{ix}=\cos(x)+i\sin(x)$, cheating about $\|z\|=1$ in order to write $$ A(x)=\frac{1}{1-z}, $$

I found that

$$ 2S(x)=\frac{\sin(x)}{1-\cos(x)} $$

as WolframAlpha does.

My question is: why with the same method I found $$ \sum_{k=0}^\infty\cos(kx)=\frac{1}{2} $$ while WolframAlpha and G.H.Hardy on the book "Divergent Series" (pg. 2) give $-1/2$?

Answer 1

Since, in general, we dont have $\lim_{k\to\infty}\sin(kx)=0$, the series $\sum_{k=0}^\infty\sin(kx)$ diverges (in other words, the sum $\sum_{k=0}^\infty\sin(kx)$ does not exist), and so does the series $\sum_{k=0}^\infty\cos(kx)$.

Answer 2

Actually:

$\sum_{k=0}^\infty\cos(kx)=\frac{1}{2}+(\frac{1}{2}+\sum_{k=1}^\infty\cos(kx))=\frac{1}{2}+\frac{1}{\pi}\delta_{0}$ in the Fourier-Stieltjes sense (as distributions for example), while

$\sum_{k=1}^\infty\sin(kx)$~$\frac{1}{2}\cot{\frac{x}{2}}=\frac{1}{2}\frac{\sin(x)}{1-\cos(x)}$ in the sense of generalized Fourier sine series

(this means that while $f$ is not necessarily in $L^1(0, \pi)$ so by odd extension it would have a unique sine series if it were integrable, $f\sin nx$ is integrable for every $n \ge 1$ and then we can compute the corresponding integrals $\pi a_n=2\int_0^{\pi}f(x)\sin nx dx$ and write $f$ ~$\sum {a_n \sin nx}$)

both quite legitimate ways of thinking about this