How to calculate$\sum\limits_{n=0}^{\infty}q^n\cos{（nθ）}$, $q\in\Bbb C$ I need judge a series is or isn't convergent,if it is convergent like that calculate the sum.The series is:$$\sum_{n=0}^{\infty}q^n\cos{(nθ)},\,|q|<{1}$$
So far,my work is:$$q^n\cos{(nθ)}$$$$=q^n\cos{(n-1)θ}\cos{θ}-q^n\sin{(n-1)θ}\sin{θ}$$$$q^n\sin{(nθ)}$$$$=q^n\cos{(n-1)θ}\sin{θ}+q^n\sin{(n-1)θ}\cos{θ}$$
Let:$$\sum_{n=0}^{\infty}q^n\cos{(nθ)}=A,$$$$\sum_{n=0}^{\infty}q^n\sin{(nθ)}=B$$So,we get:$$A=Aq\cos{θ}-Bq\sin{θ}$$$$B=Aq\sin{θ}-Bq\cos{θ}$$then:$$A\cos{θ}=Aq\cos^2{θ}-Bq\sin{θ}\cos{θ}$$$$B\sin{θ}=Aq\sin^2{θ}-Bq\cos{θ}\sin{θ}$$so:$$A\cos{θ}+B\sin{θ}=Aq$$then the A will disappear. Is my thought  wrong?
 A: You don't need to do all this to check whether a series is convergent or divergent. There is an easier method.
$|cos(\theta)|<=1$ ,$\forall \, \theta \in \mathbb R$.
and as |q|<1, the geometric progression converges.
Use the comparison test.
Hope this helps.
A: For $q\in\mathbb{R}$ we have, 
$$\sum_{n=0}^{\infty}q^n\cos{（nθ）} =\mathcal{Re(\sum_{n=0}^{\infty}q^ne^{inθ})}= \mathcal{Re(\sum_{n=0}^{\infty}z^n)} = \mathcal{Re(\frac{1}{1-z}})$$
with $z=qe^{iθ}$
If $q\in \Bbb C$ then we can write $q =|q|e^{i\alpha}$. However, $ \cos(n\theta) =\frac{e^{i\theta}+e^{-i\theta}}{2}$. 
Hence, from the previous ligne we get 
$$\sum_{n=0}^{\infty}q^n\cos{（nθ）} =\frac12\sum_{n=0}^{\infty}|q|^ne^{in(\alpha+θ)} +\frac12\sum_{n=0}^{\infty}|q|^ne^{in(\alpha-θ)} = \mathcal{Re(\sum_{n=0}^{\infty}z^n)} = \mathcal{Re}\left(\frac{1}{1-|q|e^{i(\alpha+θ)}}+\frac{1}{1-|q|e^{i(\alpha-θ)}}\right)$$
A: After observing that both $\sum_{n=0}^\infty q^n \cos(n\theta)$, $\sum_{n=0}^\infty q^n \cos(n\theta)$ are convergent (by comparison with $\sum_{n=0}^\infty \lvert q\rvert^n$), one thing you could do is to observe that $$q^n \cos(n\theta)+i q^n \sin(n\theta) = q^n e^{in\theta} = (q e^{i\theta})^n.$$
Setting $r\stackrel{\rm def}{=} q e^{i\theta}$, we have $\lvert r\rvert = \lvert q\rvert < 1$, from which
$$
\sum_{n=0}^\infty q^n \cos(n\theta) + i\sum_{n=0}^\infty q^n \sin(n\theta)
= \sum_{n=0}^\infty r^n = \frac{1}{1-r} = \frac{1}{1-qe^{i\theta}}
$$
Now, it only remains to separate the real and imaginary parts of the RHS:
$$
\sum_{n=0}^\infty q^n \cos(n\theta)
= \operatorname{Re} \left(\frac{1}{1-qe^{i\theta}}\right)
$$
(this last formula assuming $q\in\mathbb{R}$).
