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How do you calculate the infinite sum $\sum_{i=1}^{\infty} \frac{\sin(i)}{i}$? According to Wolfram Alpha, the value of the sum is $\frac{\pi - 1}{2}$, but it does not tell me the method by which it gets this result.

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Copied from this answer:

Using the power series $$ -\log(1-z)=\sum_{k=1}^\infty\frac{z^k}{k} $$ we get $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(k)}{k} &=\frac1{2i}\sum_{k=1}^\infty\frac{e^{ik}-e^{-ik}}{k}\\ &=\frac1{2i}\left[-\log(1-e^i)+\log(1-e^{-i})\right]\\ &=\frac1{2i}\log(-e^{-i})\\ &=\frac{\pi-1}{2} \end{align} $$ That is, since $1-e^{-i}$ is in the first quadrant and $1-e^i$ is in the fourth, the imaginary part of $-\log(1-e^i)+\log(1-e^{-i})$ is between $0$ and $\pi$.

Convergence is guaranteed by Dirichlet's Test and convergence to the value expected by Abel's Theorem.

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  • $\begingroup$ We are evaluating a power series on the boundary of its convergence disk, or, equivalently, we are stating that the Fourier series of the sawtooth wave is pointwise convergent at $x=1$. That is true, but it needs a bit of extra care to be justified in full rigor (this applies to my old answer in the duplicate thread, too). $\endgroup$ – Jack D'Aurizio Jan 26 '18 at 20:31
  • $\begingroup$ @JackD'Aurizio: These converge by Dirichlet's Test and converge to the value expected by Abel's Theorem. I have mentioned these in my answer. $\endgroup$ – robjohn Jan 26 '18 at 22:35
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Consider the two summations $$S=\sum_{k=1}^{\infty} \frac{\sin(k)}{k}\qquad \text{and} \qquad C=\sum_{k=1}^{\infty} \frac{\cos(k)}{k}$$ $$C+iS=\sum_{k=1}^{\infty} \frac{\cos(k)+i \sin(k)}{k}=\sum_{k=1}^{\infty} \frac{e^{i k}} k=\sum_{k=1}^{\infty} \frac{(e^{i})^ k} k=-\log \left(1-e^i\right)$$ $$C-iS=\sum_{k=1}^{\infty} \frac{\cos(k)-i \sin(k)}{k}=\sum_{k=1}^{\infty} \frac{e^{-i k}} k=\sum_{k=1}^{\infty} \frac{(e^{-i})^ k} k=-\log \left(1-e^{-i}\right)$$ Expanding the logarithm $$C+i S=-\frac{1}{2} \log \left(\sin ^2(1)+(1-\cos (1))^2\right)+i \tan ^{-1}\left(\frac{\sin (1)}{1-\cos (1)}\right)$$ $$C+i S=-\frac{1}{2} \log (2-2 \cos (1))+i \left(\frac{\pi }{2}-\frac{1}{2}\right)$$ $$C-i S=-\frac{1}{2} \log \left(\sin ^2(1)+(1-\cos (1))^2\right)-i \tan ^{-1}\left(\frac{\sin (1)}{1-\cos (1)}\right)$$ $$C-i S=-\frac{1}{2} \log (2-2 \cos (1))+i \left(\frac{1}{2}-\frac{\pi }{2}\right)$$ $$C=\frac{(C+iS)+(C-iS)}2=-\frac{1}{2} \log (2-2 \cos (1))$$ $$S=\frac{(C+iS)-(C-iS)}{2i}=\frac{1}{2} (\pi -1)$$

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