Playing with sines I wanted to find
S(x)=∞∑k=0sin(kx)
Writing it as
S(x)=Im(A(x)),whereA(x)=∞∑k=0eikx
and using z=eix=cos(x)+isin(x), cheating about ‖ 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?
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