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.

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