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This question already has an answer here:

What is the closed form of $$\sum^\infty_{n=1}\frac{\sin n}n$$?


My approach:

$$\sum^\infty_{n=1}\frac{\sin n}n=\Im\left[\sum^\infty_{n=1}\frac{e^{in}}n\right]=\Im\left[-\text{Log}(1-e^{i})\right]$$ where $\text{Log}(1):=0$.

$$\Im\left[-\text{Log}(1-e^i)\right]=-\arg(1-e^i)=-\frac{1-\pi}2$$

Thus, $$\sum^\infty_{n=1}\frac{\sin n}n=\frac{\pi-1}2$$

And as a bonus, $$\sum^\infty_{n=1}\frac{\cos n}n=\ln(2\sec1)$$


Is this correct?

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marked as duplicate by Winther, Gerry Myerson, Adrian Keister, Delta-u, Strants Sep 10 '18 at 17:25

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