A Complex Infinite Trig Sum

Let the infinite sum be:

$\begin{aligned} S(k) & = \sum_{n=1}^\infty \frac {\cos (n\theta)}{k^n} = \sum_{n=1}^\infty \frac {\Re \left(e^{in\theta}\right)}{k^n} = \Re \left(\sum_{n=1}^\infty \frac {e^{in\theta}}{k^n} \right) & \small \color{#3D99F6} \Re (\cdot) \text{ denotes the real part function.} \\ & = \Re \left(\frac {e^{i\theta}}k \cdot \frac 1{1-\frac {e^{i\theta}}k}\right) = \Re \left(\frac 1{ke^{-i\theta}-1}\right) & \small \color{#3D99F6} \text{Euler's formula: }e^{i\theta} = \cos \theta + i\sin \theta \\ & = \Re \left(\frac 1{k\cos \theta - 1 - ik\sin \theta}\right) & \small \color{#3D99F6} \text{Multiply up and down by }k\cos \theta - 1 + ik\sin \theta \\ & = \Re \left(\frac {k\cos \theta - 1 + ik\sin \theta}{k^2 - 2k\cos \theta + 1}\right) \\ & = \frac {k\cos \theta - 1}{k^2 - 2k\cos \theta + 1} & \small \color{#3D99F6} \text{Divide up and down by }k\cos \theta-1 \\ & = \frac 1{\frac {k^2-1}{k\cos \theta-1} - 2} \end{aligned}$

Therefore,

$\begin{aligned} S(2) & = S(3) \\ \implies \frac 1{S(2)} & = \frac 1{S(3)} \\ \frac {2^2-1}{2\cos \theta-1} - 2 & = \frac {3^2-1}{3\cos \theta-1} - 2 \\ \frac 3{2\cos \theta-1} & = \frac 8{3\cos \theta-1} \\ 9\cos \theta - 3 & = 16 \cos \theta - 8 \\ \implies \cos \theta & = \frac 57 \end{aligned}$

Hence $a+b = 5+7 = \boxed{12}$ .

Let $P = 1 + \frac {1}{x} \cos \theta + \frac {1}{x^2} \cos 2 \theta + \frac {1}{x^3} \cos 3 \theta...$

Let $Q = \frac {1}{x} \sin \theta + \frac {1}{x^2} \sin 2 \theta + \frac {1}{x^3} \sin 3 \theta$

(Note that these two series converge)

$P + iQ = 1 + \frac {1}{x} (\cos \theta + i\sin\theta) + \frac {1}{x^2} (\cos 2 \theta + i\sin2\theta) + \frac {1}{x^3} (\cos 3 \theta + i\sin3\theta)...$

Where $i = \sqrt{-1}$

$P + iQ = 1 + \frac {1}{x} e^{i \theta} + \frac {1}{x^2} e^{i 2\theta} + \frac {1}{x^3} e^{i 3\theta}$

This is a geometric sequence with a common ratio of $\frac {1}{x} e^{i \theta}$ .

So the converging series is equal to $\frac {1}{ 1 - \frac {1}{x}e^{i \theta}} = \frac {x}{ x -e^{i \theta}} = \frac {x}{( x -\cos \theta) - i\sin\theta}$

Rationalising the denominator we get $\frac {x^2 - x\cos\theta + ix\sin\theta}{x^2 + 1 - 2x\cos\theta}$

Taking the real part of the equation, $P = \frac {x^2 - x\cos\theta}{x^2 + 1 - 2x\cos\theta}$ .

Substituting $x = 2,3$ we get the equation $\frac {4 - 2\cos\theta}{5 - 4\cos\theta} = \frac {9 - 3\cos\theta}{10 - 6\cos\theta}$

$\Rightarrow 1 + \displaystyle \sum_{n=1}^\infty (\frac {1}{2} ^ n\cos n \theta)$ = $1 + \displaystyle \sum_{n=1}^\infty (\frac {1}{3} ^ n\cos n \theta)$

Solving this equation we know that $\cos\theta = \frac {5}{7}$ and therefore $a + b = \boxed{12}$

Using the fact $cos\theta =Re(e^{i\theta}$ ) (where $i=√(-1)$ ) and summing each side of the equality as an infinite G.P. series, we get $7cos\theta=5$ or $cos\theta=\dfrac{5}{7}$ . So $a=5, b=7$ . Hence $a+b=12$