Infinite sum

Let $\xi = \displaystyle \sum_{r=0}^\infty \frac {\cos\left(rx\right)}{b^r}$
Even though there might be other methods to solve this problem, I find Complex Numbers . the most interesting to try out.

We know, $e^{ix} = \cos x+i\sin x$

Thus,

$\xi = Re(\displaystyle \sum_{r=0}^\infty \frac {e^{irx}}{b^r})$

It is clear that this is a Geometric Progression with $\infty$ terms and with a common ratio $\frac {e^{ix}}{b}$ . Thus,

$\xi = Re(\frac {1}{1-\frac{e^{ix}}{b}})$

I leave the simplification to the reader. Therefore, we get,

$\xi = \frac {b\left(b-\cos x\right)}{b^2-2b\cos x+1}$

Using the values provided,

$\xi = \frac {6}{7}$

Thus,

$p=6$
$q=7$

And Finally,
$pq = \boxed{42}$

Let

$\displaystyle A=\sum_{r=0}^{\infty} \frac{\cos(rx)}{b^r}$

Consider

$\displaystyle B=\sum_{r=0}^{\infty} \frac{\sin(rx)}{b^r}$

We know that $e^{ix}=\cos x+i\sin x$ , hence, we have

$\displaystyle A+iB=\sum_{r=0}^{\infty} \frac{e^{irx}}{b^r}$

The RHS is an infinite geometric progression with common ratio $e^{ix}/b$ , hence

$\displaystyle A+iB=\cfrac{1}{1-\frac{e^{ix}}{b}}$

$\displaystyle \Rightarrow A+iB=\frac{b}{b-e^{ix}}=\frac{3/2}{3/2-1/2-i\sqrt{3}/2}=\frac{3}{2-\sqrt{3}i}$

Multiply and divide by $2+\sqrt{3}i$ . For the given problem, we need the real part and it comes out to be $6/7$ . Hence,

$\displaystyle p=6,q=7 \Rightarrow \boxed{pq=42}$