Given that $\rvert x \lvert \lt 1$, assuming that the series converge, prove that
$$\cos\theta+x\cos 2\theta+x^2\cos 3\theta+...+x^{r-1}\cos r\theta=\frac{\cos\theta-x}{1-2x\cos\theta+x^2}$$
I am able to prove the result but I would like to know why the series converges? How should I prove the convergence?
On
You are summing $\cos\theta+x\cos2\theta+\cdots+x^{n-1}\cos n\theta=\sum_{k=1}^{n}x^{k-1}\cos(k\theta)$.
Instead, consider the series $\sum_{k=1}^{n}x^{k-1}e^{ki\theta}=\sum_{k=1}^{n}x^{k-1}\cos(k\theta)+i\sum_{k=1}^{n}x^{k-1}\sin(k\theta)$.
Note that the series $\sum_{k=1}^{n}x^{k-1}e^{ki\theta}$ is geometric, so: $$\sum_{k=1}^{n}x^{k-1}e^{ki\theta}=\frac{1}{x}\sum_{k=1}^{n}x^{k}e^{ki\theta}=\frac{1}{x}\sum_{k=1}^{n}(xe^{i\theta})^n=\\ \frac{1}{x}\frac{xe^{i\theta}-(xe^{i\theta})^{n+1}}{1-xe^{i\theta}}. $$
Now if we try to compute the infinite sume, we have $$\sum_{k=1}^{\infty}x^{k-1}e^{ki\theta}=\frac{e^{i\theta}}{1-xe^{i\theta}}=\frac{\cos \theta+i\sin\theta}{1-x\cos\theta -ix\sin\theta}\\ =\frac{(\cos \theta+i\sin\theta)(1-x\cos\theta +ix\sin\theta)}{1-2x\cos\theta+x^2}\\ =\frac{\cos \theta-x}{1-2x\cos\theta+x^2}+\frac{i\sin\theta}{1-2x\cos\theta+x^2}=\\ \sum_{k=1}^{\infty}x^{k-1}\cos(k\theta)+\sum_{k=1}^{\infty}x^{k-1}\sin(k\theta)$$
I'm assuming it's an infinite series, otherwise convergence is not an issue.
We can show that the given series is absolutely convergent. For that observe the following: $$|\cos\theta|+|x\cos 2\theta|+|x^2\cos 3\theta| + \dotsb \leq 1+|x|+|x^2|+|x^3|+\dotsb=\frac{1}{1-|x|}.$$ The last part comes from the fact that $|x| <1$.
Since the series is AC, consequently it is convergent.