Evaluating ∑n=1∞sin⁡(nx)n without integrating ∑n=1∞enx My solution proceeds by integrating ∑n=1∞enx=eix1−eix

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Evaluating ∑n=1∞sin⁡(nx)n without integrating ∑n=1∞enx  My solution proceeds by integrating  ∑n=1∞enx=eix1−eix

Emilie Booker · Accepted Answer

Note that  ∑n=1∞sin⁡(nx)n=Im(∑n=1∞e∈xn)  =Im(∑n=1∞(eix)nn)  If x∈R {0}, then eix∈S1 {1} and therefore  ∑n=1∞(eix)nn=log⁡(11−eix)  (where log⁡ is the principal branch of the logarithm), since  log⁡(1+x)=x−x22+x33−x4x+⋯⇒log⁡(1−x)=−(x+x22+x33+x44+⋯)  ⇒log⁡(11−x)=x+x22+x33+x44+⋯  Therefore  ∑n=1∞sin⁡(nx)n=−arctan⁡(−sin⁡x1−cos⁡(x))    =arctan⁡(cot⁡(x2))   Of course, if x=0, then this equality also holds.

Evaluating ∑n=1∞sin⁡(nx)n without integrating ∑n=1∞enx My solution proceeds by integrating ∑n=1∞enx=eix1−eix

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