KVPY #19

The given series is of the form $S = \displaystyle\sum_{n=1}^{\infty} nx^{n-1}$ for $x = 1 - \cos(\theta)$ ,

where $|x| \lt 1$ since $\tan(\theta) = \sqrt{3/2}$ with $\theta$ in the first quadrant.

Now in general for $|x| \lt 1$ we have that

$R = \displaystyle \sum_{n=0}^{\infty} x^{n} = \dfrac{1}{1 - x} \Longrightarrow \dfrac{dR}{dx} = \sum_{n=1}^{\infty} nx^{n-1} = \dfrac{d}{dx}\left(\dfrac{1}{1 - x}\right) = \dfrac{1}{(1 - x)^{2}}$ ,

And so $S = \dfrac{1}{(1 - x)^{2}} = \dfrac{1}{(1 - (1 - \cos(\theta)))^{2}} = \sec^{2}(\theta) = \tan^{2}(\theta) + 1 = \dfrac{3}{2} + 1 = \dfrac{5}{2} = \boxed{2.5}$ .