# Infinite sum

$$\sum{a^ncos(nx)}$$

from zero to infinity

a is a real number -1 < a < 1

I rewrote this as a geometric series involving a complex exponential

Real part of

$$\sum{(ae^{ix})^n}$$

Which is a geometric series with common ratio r < 1, so it converges to the sum

(first term)/(1-r)

which seems to be

$$\frac{1}{1-ae^{ix}}$$

taking the real part and multiplying top and bottom by (1-acosx), I get

$$\frac{1-acos(x)}{1-2acos(x) + a^2cos^2(x))}$$

which is different from the desired result of

$$\frac{1-acos(x)}{1-2acos(x) + a^2)}$$

Any help would be appreciated

$$\sum{a^ncos(nx)}$$

from zero to infinity

a is a real number -1 < a < 1

I rewrote this as a geometric series involving a complex exponential

Real part of

$$\sum{(ae^{ix})^n}$$

Which is a geometric series with common ratio r < 1, so it converges to the sum

(first term)/(1-r)

which seems to be

$$\frac{1}{1-ae^{ix}}$$

Any help would be appreciated
From this point on this go wrong...

I know i've gone wrong somewhere. I'd like to know what I did wrong.

First work out the fraction then take the imaginary part.

Got it. Thanks.