# Thread: Convergence/Divergence of a series

1. ## Convergence/Divergence of a series

How would I determine and demonstrate whether or not the following series converges or diverges for a selected (arbitrary) real value for x:
$\sum_{n=0}^{\infty}\cos(2\pi\cdot(2^{n}x))$?
I've found clearly that it diverges for any x of the form $x=\frac{m}{2^k}$ with m and k being integers. I've also found that it diverges for $x=\frac{m}{3\cdot2^k}$. However, I don't know if it diverge for all x (as I suspect), or if there are real values of x for which the series converges. How do I find out conclusively?

--Kevin C.

2. Originally Posted by TwistedOne151
How would I determine and demonstrate whether or not the following series converges or diverges for a selected (arbitrary) real value for x:
$\sum_{n=0}^{\infty}\cos(2\pi\cdot(2^{n}x))$?
I've found clearly that it diverges for any x of the form $x=\frac{m}{2^k}$ with m and k being integers. I've also found that it diverges for $x=\frac{m}{3\cdot2^k}$. However, I don't know if it diverge for all x (as I suspect), or if there are real values of x for which the series converges. How do I find out conclusively?

--Kevin C.
Off the top of my head, perhaps a proof along the following lines:

Using the double angle formula:

$\cos(2\pi\cdot(2^{n}x) = \cos(2[\pi\cdot(2^{n}x)]) = 1 - 2 \sin^2(\pi\cdot(2^{n}x))$.

The sum of 1's certainly diverges. If the sum of $sin^2$ 's converges then overall you've got divergence ..... On the other hand, if twice the sum of $sin^2$ 's diverges, it'll diverge much more slowly than the sum of 1's and so you've got divergence again ....

Sloppy I know (and the 'other hand' claim might prove to be bogus) - over to you (or others) to dot the i's and cross the t's (and put in the missing words and supply the punctuation)

3. Originally Posted by TwistedOne151
How would I determine and demonstrate whether or not the following series converges or diverges for a selected (arbitrary) real value for x:
$\sum_{n=0}^{\infty}\cos(2\pi\cdot(2^{n}x))$?
I've found clearly that it diverges for any x of the form $x=\frac{m}{2^k}$ with m and k being integers. I've also found that it diverges for $x=\frac{m}{3\cdot2^k}$. However, I don't know if it diverge for all x (as I suspect), or if there are real values of x for which the series converges. How do I find out conclusively?

--Kevin C.
Or, of course, an even simpler proof that applies the limit test (also known as the nth term test) .......

Let $a_n = cos(2\pi\cdot(2^{n}x))$. Since the sequence $\{a_n\}$ doesn't converge to zero, the series is divergent.