Question

Find those values of x satisfying 0 <= x <= 2pi for which the geometrical series:
1 + 2cosx + 4cos^2x + 8cos^3x + ... has a limiting sum.

I'm wondering how we'd approach this question?
Maybe by sketching y= cos x for 0 <= x <= 2pi ?? :)

Answer 1

Start by finding the common ratio

Answer 2

common ratio = 2cosx

Answer 3

whats the criterion for geometric series to converge ?

Answer 4

common ratio must be between -1 and 1, yes ?

Answer 5

yesss

Answer 6

-1 < 2cosx < 1

solve x

Answer 7

ohh okay
so,
\[\frac{ \pi }{ 3 } < x < \frac{ 2\pi}{ 3 } and \frac{ 4\pi }{ 3 } < x < \frac{ 5\pi }{ 3 }\]

Answer 8

looks good
http://www.wolframalpha.com/input/?i=solve+%7C2cos%28x%29%7C%3C1%2C+0%3Cx%3C2pi

Answer 9

thank you :)
I forgot about the criterion for geometric series LOL.