A Harmonic Hex

Let $(x,y)$ be the coordinates of $P$ . It is easy to see that $x=\sum_{n=0}^{\infty}\frac{\cos{n\theta}}{n+1}$ $y=\sum_{n=0}^{\infty}\frac{\sin{n\theta}}{n+1}$ where $\theta=\frac{\pi}{3}$ . Instead of trying to find these sums directly, let's exploit complex numbers. Consider $x+iy=\sum_{n=0}^{\infty}\frac{\cos{n\theta}+i\sin{n\theta}}{n+1}=\sum_{n=0}^{\infty}\frac{e^{in\theta}}{n+1}$ Let $z=e^{i\theta}$ , then we have $\sum_{n=0}^{\infty}\frac{z^{n}}{n+1}=-\frac{\ln{(1-z)}}{z}$ That last step comes from the Taylor expansion for $\ln{(1+x)}$ about $x=0$ . Note that $z=\cos{\theta}+i\sin{\theta}=\frac{1}{2}+i\frac{\sqrt{3}}{2}$ .
Now, $-\frac{\ln{(1-z)}}{z}=-\overline{z} \space \ln{(\frac{1}{2}-i\frac{\sqrt{3}}{2})}=-\overline{z} \space \ln{e^{-i\frac{\pi}{3}}}$ $=i \space \overline{z}\frac{\pi}{3}=x+iy$ Taking the modulus of both sides, and noting that $i \space \overline{z}$ has modulus 1 we arrive at $\sqrt{x^2+y^2}=OP=\boxed{\frac{\pi}{3}}$ .

we can do this problem by taking a segment as a vector or complex number we know that the sum of all such complex numbers of vectors will give by vector addition the vector joining origin and the approaching point. Any general segment when moved parallel so that it passes through the origin with same magnitude can be represented by

$Z=\frac{1}{r}e^{i(r-1)\pi/3}$ so the addition of all such complexes is given by $\sum_{1}^{\infty}Z=\frac{1}{r}e^{i(r-1)\pi/3}=\frac{\pi}{3}$

Let the $n$ th segment be represented by a complex number $u_n$ . For example, $u_1 = 1$ , $u_2 = \frac 12 e^{i\frac \pi 3}$ , $u_3 = \frac 13 e^{i\frac {2\pi}3}$ , ... $u_n = \frac 1n e^{i\frac {n\pi}3}$ . Then the coordinates of the end of the $n$ th segment is given by $P_n (\Re(z_n), \Im(z_n))$ , where $z_n = \sum_{k=1}^n \frac 1k e^{i\frac {(k-1)\pi}3}$ . Then the distance from the origin $OP_n = |z_n|$ . Then we have:

$\begin{aligned} \lim_{n \to \infty} z_n & = \sum_{k=1}^\infty \frac {e^{i\frac {(k-1)\pi}3}}k = e^{-i\frac \pi 3} \sum_{k=1}^\infty \frac {e^{i\frac {k\pi}3}}k = e^{-i\frac \pi 3} \ln \left(1-e^{i\frac \pi 3} \right) \\ & = e^{-i\frac \pi 3} \ln \left(1-\frac 12 - i\frac {\sqrt 3}2 \right) = - e^{-i\frac \pi 3} \ln \left(\frac 12 - i\frac {\sqrt 3}2 \right) \\ & = - e^{-i\frac \pi 3} \ln \left(e^{-i\frac \pi 3} \right) = i \frac \pi 3 e^{-i\frac \pi 3} \end{aligned}$

Therefore, $\displaystyle OP = \left| \lim_{n \to \infty} z_n \right| = \left| i \frac \pi 3 e^{-i\frac \pi 3} \right| = \frac \pi 3 \implies a + b = 1+3 = \boxed 4$ .

The position of P as a complex number can be written as :

1 + e^(i pi/3)/2 + e^(2 i pi/3)/3 +...... which is of the form sum (from n = 1 to infinity ) [x^n / n] (where x = e^i pi/3. ) = sum (from n = 1 to infinity ) [x $x^(n-1)dx] = x $ 1 dx / (1-x) (bringing the infinite series summation inside = - x ln | 1 -x | = - e^(i pi/3) ln | 1 - e^(i pi/3) | Using eulers formula and separating the real and imaginary parts, we get x coordinate of P = pi/6, y coordinate of P = sqrt(3) pi / 6

Using euclidean distance formula, the distance of p from the origin = sqrt [ pi^2/36 + 3pi^2/36] = sqrt[4pi^2/36] = pi /3, thus giving 4 as the required answer