Answer: 36

Small Hint:

Combine the two series as $Q + iP:$ the terms become a geometric series with ratio $\frac{ie^{i\theta}}{2}$

Big Hint:

Summing gives $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$ Set it equal to $\frac{2\sqrt{2}}{7},$ square, and keep the root with $\sin\theta \le 0.$

Solution:

The signs and the alternation between sines and cosines suggest powers of $i:$ indeed $$\begin{aligned} Q + iP &= 1 + \frac{1}{2}ie^{i\theta} + \frac{1}{4}i^2e^{2i\theta} \\ &\quad {}+ \frac{1}{8}i^3e^{3i\theta} + \cdots \\ &= \frac{1}{1 - \frac{ie^{i\theta}}{2}} = \frac{2}{2 - ie^{i\theta}}. \end{aligned}$$ Since $2 - ie^{i\theta} = (2 + \sin\theta) - i\cos\theta,$ multiplying by the conjugate gives $$\begin{aligned} &Q + iP \\ &= \frac{2(2 + \sin\theta) + 2i\cos\theta}{5 + 4\sin\theta}, \end{aligned}$$ so $\frac{P}{Q} = \frac{\cos\theta}{2 + \sin\theta}.$

Setting $\frac{\cos\theta}{2 + \sin\theta} = \frac{2\sqrt{2}}{7}$ and squaring, $49(1 - \sin^2\theta) = 8(2 + \sin\theta)^2,$ which rearranges to $$\begin{aligned} &57\sin^2\theta + 32\sin\theta - 17 \\ &= (3\sin\theta - 1)(19\sin\theta + 17) \\ &= 0. \end{aligned}$$

Since $\pi \le \theta \lt 2\pi$ forces $\sin\theta \le 0,$ we get $\sin\theta = -\frac{17}{19}$ (and then $\cos\theta = \frac{6\sqrt{2}}{19} \gt 0,$ consistent with the positive ratio). Thus $m + n = 17 + 19 = 36.$