Geometric progression of 1 and 1/3tan^2θ **The first two terms of a geometric progression are  where $0<θ<π/2$.
(i) Find the set of values of θ for which the progression is convergent. [2]**
What does convergent mean and how to solve this?
 A: Hint: The infinite geometric series  $1+x+x^2+x^3+\cdots$ is convergent if and only if $|x|\lt 1$.
Remark: As for the definition of convergence, let $a_0+a_1+a_2+\cdots$ be an infinite series. 
For any $n$, let $s_n$ be the partial sum $a_0+a_1+\cdots +a_n$. We say that the series $a_0+a_1+a_2+\cdots$ converges if the sequence $(s_n)$ of partial sums has a (finite) limit. 
A: An infinite geometric progression with initial term $1$ and common ratio $r$ can be denoted by:
$$1, r, r^2, r^3,..., r^n, ...$$
For the progression to be convergent, the limit $\lim_{n \to \infty} r_n$ ("final term") has to be zero, meaning that $|r| < 1$.
So in this case, $r = \frac{1}{3}\tan^2 \theta$. This is always non-negative so you don't really have to worry about the absolute value. All you need to do is solve the inequality:
$$\frac{1}{3}\tan^2 \theta < 1$$
within the given range of values for $\theta$.
Solving,
$$-\sqrt 3 < \tan \theta < \sqrt3$$
Within the required range, that just leaves $0 <\theta < \frac{\pi}{3}$, since the tangent is a strictly increasing function over the relevant domain.
