Thursday, 18 October 2018

sequences and series - Evaluating sumik=0nftysin(kx) and sumik=0nftycos(kx)



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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