# Convergence of series of the form (r^n)cos(nx)

1. Aug 31, 2008

### buffordboy23

Show that

$$\sum^{\infty}_{n=0} p^{n}cos\left(n\theta\right) = \frac{1-pcos\theta}{1-2pcos\theta+p^{2}}$$

if $$\left|p\right|<1$$.

Looking at this series, I see that p will approach zero as n approaches infinity, while the series oscillates because of the cosine term. The convergence is easy to see, but not the function that it converges to. The series is reminiscent of a geometric series, or possibly a Fourier series, but the cosine term really complicates things for me. I tried looking for similar problems on the web, but only found examples that apply the tests of convergence rather than determining this function.

2. Aug 31, 2008

### tiny-tim

Hi buffordboy23!

erm … if p is real, then $$p^{n}cos\left(n\theta\right) = \Re(p^ne^{in\theta})$$

3. Aug 31, 2008

### buffordboy23

tiny-tim,

I am still clueless. I have thought of this option but still cannot proceed. I have tried various expansions and equivalent expressions for the initial cosine term--such as cos nx = [exp(nx) + exp(-nx)]/2--and the equivalent form that you suggest. Unfortunately, my text is rather terse on the convergence of an infinite series, as it only discusses the Cauchy-criterion and Weierstrauss condition--methods to show convergence, but not the function that it is convergent to--and the text is also devoid of example problems. Is there a specific name of the concept to be used in this application?

4. Aug 31, 2008

### tiny-tim

Hi buffordboy23!

Can you sum $$\sum_{n = 0}^\infty p^n$$ (for |p| < 1) ?

5. Aug 31, 2008

### buffordboy23

Yes, this is just a geometric series, where the constant term is 1, so the sum of this series is

$$\frac{1}{1-p}$$

EDIT: I never saw anything before with the form cos(nx), and could not find any examples anywhere either.

6. Aug 31, 2008

### tiny-tim

ok … now apply that to $$\sum_{n=0}^\infty p^ne^{in\theta}$$

7. Aug 31, 2008

### buffordboy23

I am still confused; I don't ever recall seeing the summation of a complex exponential. I am thinking that I can somehow split the terms so that I have two summation terms, which then will lead to the product of two separate terms.

8. Aug 31, 2008

### tiny-tim

$$\sum_{n=0}^\infty p^ne^{in\theta}\ =\ \sum_{n=0}^\infty \left(pe^{i\theta}\right)^n$$

9. Aug 31, 2008

### buffordboy23

Yes...I see it clearly now. The funny thing is that I had this form at one moment, but failed to recognize its obscure relation to the geometric series.

Thanks tiny-tim for your time and assistance.