Apple sum

$\begin{aligned} \sum_{n=1}^\infty \frac {\sin (n)}n & = \blue \Im \left(\sum_{n=1}^\infty \frac \blue{e^{ni}}n \right) & \small \blue{\text{Since }e^{\theta i} = \cos \theta + i \sin \theta \text{ and }\Im(\cdot)} \\ & = \Im \left(-\ln \left(1-e^i \right) \right) & \small \blue{\text{is the imaginary part function.}} \\ & = - \Im \left(\ln \left(1-\cos 1 - i\sin 1 \right) \right) \\ & = - \Im \left(\ln \left(\sqrt{(1-\cos 1)^2 + \sin^2 1} \cdot e^{\tan^{-1} \left(\frac {\sin 1}{1-\cos 1}\right) i} \right) \right) \\ & = - \Im \left(\ln \left(\sqrt{(1-\cos 1)^2 + \sin^2 1}\right) - i \tan^{-1} \left(\frac {\sin 1}{1-\cos 1}\right) \right) \\ & = \tan^{-1} \left(\frac {\sin 1}{1-\cos 1} \right) = \tan^{-1} \left(\frac {\frac {2t}{1+t^2}}{1-\frac {1-t^2}{1+t^2}} \right) = \tan^{-1} \left(\frac 1t \right) & \small \blue{\text{Let }t = \tan \frac 12} \\ & = \tan^{-1} \left(\frac 1{\tan \frac 12} \right) = \frac 12 (\pi -1) \approx \boxed{1.07} \end{aligned}$

$\textrm{Apple sum 🍎🍏}$

$\sum_{n=1}^∞ \frac{\sin(n)}{n}$

$= \sum_{n=1}^∞ \frac{e^{in} -e^{-in}}{2in}$

$\implies{\frac{1}{2i}(\sum_{n=1}^∞ \frac{e^{in}}{n} - \frac{e^{-in}}{n})}$

$\implies{\frac{1}{2i}(- log(1-e^i )+log(1-e^{-i}))}$

So, $=\frac{1}{2i} log(\frac{1-e^{-i}}{1-e^i})$

$= \frac{1}{2i} log(-e^{-i}) = \frac{1}{2i} (πi-i)$

Answer is $\boxed{\frac{π-1}{2}} :) 🍎🍏$