Curious Series

By the Dirichlet test for convergence we can anticipate that the series does converge.

We'll look first in general at $S(x) = \displaystyle\sum_{k=1}^{\infty} \dfrac{\sin(kx)}{k},$ and then evaluate $S(\frac{\pi}{180}).$

Now we have that $\sin(y) = \dfrac{e^{iy} - e^{-iy}}{2i},$ so

$S(x) = \displaystyle\sum_{k=1}^{\infty} \dfrac{e^{ikx} - e^{-ikx}}{2ki} = \dfrac{-\ln(1 - e^{ix}) + \ln(1 - e^{-ix})}{2i},$

since in general $\displaystyle\sum_{k=1}^{\infty} \dfrac{z^{k}}{k} = -\ln(1 - z).$

(I'm playing a bit fast and loose with radius of convergence in $\mathbb{C}$ here; I'll try and deal with that later.)

Continuing, we can then say that

$S(x) = \dfrac{1}{2i}\ln\left(\dfrac{1 - e^{-ix}}{1 - e^{ix}}\right) = \dfrac{1}{2i}\ln(-e^{-ix}) = \dfrac{1}{2i}\ln(e^{i(\pi - x)}) = \dfrac{i(\pi - x)}{2i} = \dfrac{\pi - x}{2},$

since $\dfrac{-e^{-ix}}{-e^{-ix}} * \dfrac{1 - e^{-ix}}{1 - e^{ix}} = \dfrac{-e^{-ix}(1 - e^{-ix})}{1 - e^{-ix}} = -e^{-ix},$ and since

$-e^{-ix} = -\cos(x) + i*\sin(x) = \cos(\pi - x) + i*\sin(\pi - x) = e^{i(\pi - x)}.$

Finally, $S(\frac{\pi}{180}) = \dfrac{\pi - \frac{\pi}{180}}{2} = \dfrac{179 \pi}{360},$ and thus $a + b = 179 + 360 = \boxed{539}.$

I like Brian's solution! However, Brian is rightly concerned about convergence issues; perhaps we can avoid those problems by considering a power series.

Let's start with the geometric series $\sum_{k=1}^{\infty}e^{ika}x^k=\frac{e^{ia}x}{1-e^{ia}x}$ where $0<x<1$ . Now we can divide by $x$ and integrate to find that $\sum_{k=1}^{\infty}\frac{e^{ika}}{k}x^k=-\ln(1-e^{ia}x)$ We can take the limit as $x$ approaches 1 from below and use Abel's Theorem to see that $\sum_{k=1}^{\infty}\frac{e^{ika}}{k}=-\ln(1-e^{ia})$ Considering the imaginary parts, we end up with $\sum_{k=1}^{\infty}\frac{\sin(ka)}{k}=-\arg(1-e^{ia})$ If $a$ is between 0 and $2\pi$ , the RHS is $\frac{\pi-a}{2}$ ; perhaps the fastest way to see this is to draw the isosceles triangle with its vertices at $0,1$ and $e^{ia}$ in the Argand plane. For $a=\frac{\pi}{180}$ the sum is $\frac{179\pi}{360}$ and the required answer is $\boxed{539}$

Let us define, for $z\in \mathbb{R}$ , the series $A(z)=\sum_{k\ge 1}\frac{z^k\sin (ak)}{k},\ a=\frac{\pi}{180}$ We can see that for $z\in (-1,1)$ the series is absolutely summable and convergent. Also, I think by Monotone convergence theorem, if follows that $\lim_{z\to 1}A(z)=\sum_{k\ge 1}\frac{\sin(ak)}{k}$ though it might need some care. Assuming it is correct, now we can proceed to find an expression for $A(z)$ . Note that $z\in (-1,1)\implies$ $A(z)=\Im\sum_{k\ge 1}\frac{z^k\exp(iak)}{k}\\\implies \frac{dA}{dz}=\Im\sum_{k\ge 1}z^{k-1}\exp(iak)\\=\frac{1}{z}\Im\sum_{k\ge 1}\frac{z}{\exp(-iak)-z}\\=\frac{\sin a}{(z-\cos a)^2+\sin^2 a}\\\implies A(z)=\arctan\left(\frac{z-\cos a}{\sin a}\right)+C$ Now, $A(0)=0\implies C=\arctan\cot(a)=\frac{\pi}{2}-a$ . Hence $A(z)=\arctan\left(\frac{z-\cos a}{\sin a}\right)+\frac{88\pi}{180}\implies \lim_{z\to 1}A(z)=\arctan\left(\frac{1-\cos a}{\sin a}\right)+\frac{88\pi}{180}\\=\frac{\pi}{360}+\frac{88\pi}{180}=\frac{179\pi}{360}$ So, $\boxed{539}$ is the answer.

Very nice series problem, Otto! Summing an infinite discretized "sinc" pulse.........nice little midnight signal processing problem for EE's!