sequences and series - Why does $sum_{k=1}^{infty}frac{{sin(k)}}{k}={frac{pi-1}{2}}$?

Inspired by this question (and far more straightforward, I am guessing), Mathematica tells us that $$\sum_{k=1}^{\infty}\dfrac{{\sin(k)}}{k}$$ converges to $\dfrac{\pi-1}{2}$.

Presumably, this can be derived from the similarity of the Leibniz expansion of $\pi$ $$4\sum_{k=1}^{\infty}\dfrac{(-1)^{k+1}}{2k-1}$$ to the expansion of $\sin(x)$ as $$\sum_{k=0}^{\infty}\dfrac{(-1)^k}{(2k+1)!}x^{2k+1},$$ but I can't see how...

Could someone please explain how $\dfrac{\pi-1}{2}$ is arrived at?

Answer

Here is one way, but it does not use the series you mention so much. I hope that's OK.

The series is:

$$\sin(1)+\frac{\sin(2)}{2}+\frac{\sin(3)}{3}+\cdot\cdot\cdot$$

$$\Im\left[e^{i}+\frac{e^{2i}}{2}+\frac{e^{3i}}{3}+\cdot\cdot\cdot \right]$$

Let $\displaystyle x=e^{i}$.

$$\Im\left[x+\frac{x^2}{2}+\frac{x^3}{3}+\cdot\cdot\cdot \right]$$

differentiate:

$$\Im \left[1+x+x^{2}+x^{3}+\cdot\cdot\cdot \right]$$

This is a geometric series, $\displaystyle \frac{1}{1-x}$

$$\Im [\frac{1}{1-x}]$$

Integrate:

$$-\Im[\ln(x-1)]=-\Im [\ln(e^{i}-1)]$$

Now, suppose $$\ln(e^{i}-1)=a+bi$$ ,

$$e^{i}-1=e^{a}e^{bi}$$

$$\cos(1)-1+i\sin(1)=e^{a}\left[\cos(b)+i\sin(b)\right]$$

Equate real and imaginary parts:

$$\cos(1)-1=e^{a}\cos(b)\\ \sin(1)=e^{a}\sin(b)$$

divide both:

$$\frac{\cos(1)-1}{\sin(1)}=\frac{e^{a}\sin(b)}{e^{a}\cos(b)}$$

$$-\cot(1/2)=\tan(b)$$

$$b=\tan^{-1}(\cot(1/2))=\frac{1}{2}-\frac{\pi}{2}$$

But we need the negative of this, so finally:

$$\frac{\pi}{2}-\frac{1}{2}$$