Harmonious sign

Use Euler's formula to get that : $L=\sum_{k=1}^{\infty} \frac{\sin k }{k} =\text{Im} \sum_{k=1}^{\infty} \frac{e^{ik}}{k} = \text{Im}(-\ln(1-e^i)).$ Now, note that : $1-e^i =1-\cos 1 -\sin 1 = 2 \sin \frac{1}{2} \left(\sin \frac{1}{2}-\cos \frac{1}{2}\right)=2\sin \frac{1}{2} e^{\frac{1-\pi}{2}}.$ Therefore, the wanted limit is : $\text{Im}(-\ln(1-e^i))= \frac{\pi-1}{2} \approx1.071.$

I took the Fourier Transform of the rectangle function with width $2a$ , centered at the origin. Then I determined that taking $a=\frac{1}{2\pi}$ makes the result $\frac{\sin{s}}{\pi s}$ . From there I could apply the Poisson summation formula to conclude that the sum of the rectangle function with width $\frac{1}{2\pi}$ , centered at the origin, was equal to the sum of $\frac{\sin{s}}{\pi s}$ over all the integers (including 0 and negative integers). The sum of the rectangle function over the integers is equal to 1 because it takes the value 1 at 0, but is 0 for all the other integers. Thus, $\sum_{k=-\infty}^\infty \frac{\sin{s}}{\pi s} = 1 \Rightarrow \sum_{k=-\infty}^\infty \frac{\sin{s}}{s} = \pi \Rightarrow \sum_{k\neq 0} \frac{\sin{s}}{s} = \pi -1 \Rightarrow \sum_{k=1}^\infty \frac{\sin{s}}{s} = \frac{\pi-1}{2}$ .

Essentially, the series is the fourier series of the function f(x) = x/2 from 0 to 2*pi resulting in a series of the form

pi/2 - (sin(x) + sin(2x) / 2 + .....sin(ix) / i + .......) the series in the brackets being the series that we want to evaluate

If we substitute x = 1 , we get the given series So we get 1/2 = pi/2 - S (S being the given series)

So, S = pi/2 - 1/2