An expression for $\large \frac{1}{\pi}$

So I cooked up this expression for $1/\pi$, but I'm not immediately able to prove it, although computation shows it to be true. Perhaps somebody can think of a neat proof!

$$\mathrm{Let}~~S_n=\{\omega \in \mathbb{C}: w^n=1\}.\quad \mathrm{Let}~~ a_n=\frac{1}{n}~\sup_{T~ \subseteq S_n}\left|\sum_{\omega \in T}\omega \right|.$$

$$\mathrm{Then}~~ a_n \to \frac{1}{\pi} ~\mathrm{as}~~ n \to \infty.$$

(It seems that the value of $a_n$ can usually be attained by picking all of those $n^{th}$ roots of unity in the upper half plane, for example. The limit would follow from a proof of this observation.)

Thanks, and enjoy!

A sketch. Consider a given $T$ and the corresponding sum $S$ as a vector in the complex plane. Let $L$ be its perpendicular through the origin. If there are any elements of $T$ on the wrong side of $L$, omitting them increases the length of $S$, so WLOG every element of $T$ lies in the upper half plane. It's not hard to see that if $T$ isn't maximal, adding another element of $S_n$ in the upper half plane increases the length of $S$, so the conclusion follows.