# Ask Uncle Colin: A Geometric Sequence with a Cosine in it

Dear Uncle Colin,

The first term of a geometric progression is 1 and the second is $2\cos(x)$, with $0 \lt x \le \piby 2$. Find the set of values for which the progression converges.

Read And Typed It Out

Hi, RATIO, and thanks for your message! It looks a bit like you’re asking me to do your homework, which is why I’ve waited so long to respond. Most AUC correspondents at least *pretend* to be interested in the answer.

For the benefit of other people, here’s how I’d tackle it.

The key thing to a geometric progression converging is that the common ratio is between -1 and 1 – strictly((That means: it can’t **be** -1 or 1, but it can be as close as you like on the correct side.)).

Here, we know the first term is $a=1$ and the second is $ar = 2\cos(x)$, so $r = 2\cos(x)$.

Therefore, we need $-1 < 2\cos(x) < 1$ for the sequence to converge – or, put another way, $-\frac{1}{2} < \cos(x) < \frac{1}{2}$.

I would draw a graph of $y = \cos(x)$ for $0 < x < \piby2$, starting flat at $(0,1)$ and gradually getting steeper as it descends to $\left(\piby2, 0\right)$.

It reaches $y=\frac{1}{2}$ at $x = \piby 3$, and – taking adequate care with the strictness of the inequality – we can write down an answer of $\piby3 < x < \piby 2$.

Hope that helps… somebody!

- Uncle Colin