Find the set of values of $\theta$ lying in the interval $-\dfrac{\pi}{2}\!<\!\theta\!<\!\dfrac{\pi}{2}$ such that the sum to infinity of the geometric series

1. Identify the type of series and state the condition for convergence.

The given series is a geometric progression with first term $a = 1$ and common ratio $r = \tan\theta$.
For the sum to infinity to exist, we must have $|r| < 1$:
$|\tan\theta| < 1$
$-1 < \tan\theta < 1$

2. Formulate the expression for the sum to infinity.

The sum to infinity is given by $S_\infty = \frac{a}{1-r}$.
Substituting our values:
$S_\infty = \frac{1}{1 - \tan\theta}$

3. Set up the given inequality.

We are given that the sum to infinity is less than $\frac{3+\sqrt{3}}{2}$:
$\frac{1}{1 - \tan\theta} < \frac{3+\sqrt{3}}{2}$
From the convergence condition $-1 < \tan\theta < 1$, we know that $1 - \tan\theta > 0$.
Thus, we can multiply both sides by $(1 - \tan\theta)$ without reversing the inequality sign:
$1 < \left( \frac{3+\sqrt{3}}{2} \right) (1 - \tan\theta)$

4. Solve the inequality for $\tan\theta$.

Rearranging the inequality to solve for $\tan\theta$:
$1 - \tan\theta > \frac{2}{3+\sqrt{3}}$
Rationalising the denominator:
$1 - \tan\theta > \frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}$
$1 - \tan\theta > \frac{6-2\sqrt{3}}{9-3}$
$1 - \tan\theta > \frac{6-2\sqrt{3}}{6}$
$1 - \tan\theta > 1 - \frac{\sqrt{3}}{3}$
$-\tan\theta > -\frac{\sqrt{3}}{3}$
$\tan\theta < \frac{\sqrt{3}}{3}$

5. Combine the conditions to find the exact range of values for $\theta$.

To satisfy both the existence of the sum and the given inequality, we combine the conditions:
$-1 < \tan\theta < \frac{\sqrt{3}}{3}$
Given the interval $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, and knowing $\tan^{-1}(-1) = -\frac{\pi}{4}$ and $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$, the set of values for $\theta$ is:
$-\frac{\pi}{4} < \theta < \frac{\pi}{6}$