Infinite Cosine Series

Let $S =\displaystyle \sum _{k=1} ^{\infty} \dfrac{k\cos(k\theta)}{5^k}$ .

Integrating $S$ , in terms of $\theta$ , leads us to:

$\int S d\theta =\displaystyle \sum _{k=1} ^{\infty} \dfrac{\sin(k\theta)}{5^k} = \dfrac{1}{2i} \sum _{k=1} ^{\infty} \left[ \left( \dfrac{e^{i\theta}}{5} \right)^k -\left( \dfrac{e^{-i\theta}}{5} \right)^k \right] = \dfrac{1}{2i} \left( \dfrac{\dfrac{e^{i\theta}}{5}}{1 - \dfrac{e^{i\theta}}{5}} - \dfrac{\dfrac{e^{-i\theta}}{5}}{1 - \dfrac{e^{-i\theta}}{5}} \right)$

(The last term above was given by the infinite geometric series' formula.)

Rearranging the last term above, we have:

$\int S d\theta = \dfrac{1}{2i} \left( \dfrac{\dfrac{e^{i\theta}}{5}}{1 - \dfrac{e^{i\theta}}{5}} - \dfrac{\dfrac{e^{-i\theta}}{5}}{1 - \dfrac{e^{-i\theta}}{5}} \right) = \dfrac{5\sin(\theta)}{26 - 10\cos(\theta)}$

Differentiating both sides:

$S = \dfrac{5(13\cos(\theta) - 5)}{2(13-5\cos(\theta))^2}$

Given that $\cos(\theta) = \dfrac{7}{10}$ , then $S = \dfrac{m}{n} = \dfrac{41}{361}.$

Finally, $m+n = 41 + 361 = \boxed{402}$

Let $S=\displaystyle \sum_{n=1}^{\infty} \dfrac{n \cos(n\theta)}{5^n}$ . Now, let $w=\displaystyle \sum_{n=1}^{\infty} \dfrac{n (e^{i\theta})^n}{5^n}$ . Then $S=\text{Re}(w)$ . Let's find $w$ :

$w=\displaystyle \sum_{n=1}^{\infty} n\left(\dfrac{e^{i\theta}}{5}\right)^n$

Use the formula for an arithmetic-geometric sum: $\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$ for $|x|<1$

$w=\dfrac{\dfrac{e^{i\theta}}{5}}{\left(1-\dfrac{e^{i\theta}}{5}\right)^2}=\dfrac{5e^{i\theta}}{(5-e^{i\theta})^2}=\dfrac{5e^{i\theta}}{e^{i\theta}(5e^{-i\theta/2}-e^{i\theta/2})^2}=\dfrac{5}{25e^{-i\theta}-10+e^{i\theta}}$

$w=\dfrac{5}{26\cos(\theta)-10-24i\sin(\theta)}=\dfrac{5(13\cos(\theta)-5+12i\sin(\theta))}{2((13\cos(\theta)-5)^2+(12\sin(\theta))^2)}=\dfrac{5(13\cos(\theta)-5+12i\sin(\theta))}{2(13-5\cos(\theta))^2}$

Finally, $S=\text{Re}(w)=\dfrac{5(13\cos(\theta)-5)}{2(13-5\cos(\theta))^2}$ .

Substituting $\cos(\theta)=\dfrac{7}{10}$ we get $S=\dfrac{41}{361}$ . So the answer is $41+361=\boxed{402}$ .

$S(x) = x(1-x)^{-2} = x+2x^2+3x^3+...$
So the required sum will be real part of $S(\frac{e^{i\theta}}{5})={41 \over 361}, \cos \theta = \frac{7}{10}$ .