sequences and series - Show that $sum_{k=0}^{infty} p^kcos(kx) = frac{1-pcos(x)}{1-2pcos(x)+p^2}$ using complex numbers.

I have this equality that I'm trying to show. I have tried many times but I can't get it to work.

Could you please help me?

I would like to first make this finite sum equal something.

$$\sum_{k=0}^np^k\cos(kx) = S_{n}(p, x), -1 < p < 1 $$

And then use it to calculate the infinite sum, which should be the right side.

$$\lim_{n \to {\infty}} S_n(p, x)=\sum_{k=0}^{\infty}p^k\cos(kx) = \frac{1-p\cos(x)}{1-2p\cos(x)+p^2}$$

I would like to use complex numbers to show this.

EDIT: I know that I have to use Euler's identity, I just can't get the algebra to work.

I have gotten so far but I don't know how to continue.

$$\frac{(pe^{ix})^{(n+1)/2}((pe^{ix})^{-(n+1)/2}-(pe^{ix})^{(n+1)/2})}{(pe^{ix})^{1/2}((pe^{ix})^{-1/2}-(pe^{ix})^{1/2})}$$

Thank you in advance!

Answer

If $p$ is a real number then $ p^k \cos(kx)$ is the real part of the complex number
$$
p^k (\cos(kx) + i \sin(kx)) = p^k e^{ikx} = (pe^{ix})^k \, .
$$
So you can compute the geometric sum
$$

\sum_{k=0}^n (pe^{ix})^k = \frac{1-(pe^{ix})^{n+1}}{1-pe^{ix}}
$$
and determine its real part. That is done by expanding the fraction with the conjugate of the denominator:
$$
\frac{(1-(pe^{ix})^{n+1})(1-pe^{-ix})}{(1-pe^{ix})(1-pe^{-ix})}
= \frac{1 -p^{n+1} e^{i(n+1)x} -pe^{-ix} + p^{n+2}e^{inx} }{1-2p \cos(x) + p^2} \, .
$$
It follows that
$$
\sum_{k=0}^np^k\cos(kx) = \frac{1 - p^{n+1}\cos((n+1)x) - p \cos(x) + p^{n+2} \cos(nx)}{1-2p \cos(x) + p^2} \, .

$$