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Evaluate $\sum_{n=1}^\infty \dfrac{\sin n}n$.

By Euler's theorem, $\sin x = \dfrac{e^{ix}-e^{-ix}}{2i}$ for every real number x. Also, I know that $\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}6.$ I'm not sure whether the integral $\int_{1}^\infty \dfrac{\sin x}x dx$ converges. One method for computing the exact value of an infinite sum is to use the Squeeze theorem, but I'm not sure how to do that here since the partial sums of the given series likely do not have a closed form. Another method is telescoping series, though it's unclear whether one can even find such a series in this case.

As a side note, the following link seems to provide useful info about the complex logarithm: https://www.maths.tcd.ie/~dwilkins/Courses/214/214S2_0708.pdf.

user3379
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  • As you said, use Euler's theorem, and recall the logarithmic series $\log(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}x^n$. – Nicolas Bourbaki Sep 24 '22 at 01:58
  • @NicolasBourbaki in the answer to the duplicate by Jeb, what do they mean by "choose the right branch"? – user3379 Sep 24 '22 at 03:01
  • It means that when you work with the complex-logarithm, $\log(z)$ might have different meanings to it, you need to choose the right "branch" as in the correct "root" out of many. – Nicolas Bourbaki Sep 24 '22 at 05:40

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