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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!

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    @amWhy: A semantic quibble: `\mathrm` is intended for setting *math* upright (as in $\det{A}$), while `\text` is the command to write text in a math environment. Also, you can get the spacing right by inserting it into the text environment `\text{Let } S_n` instead of tildes.2011-07-08

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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.

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    Cool! If anyone feels like writing up a complete solution, I'll accept it as the answer. If not, I'll accept Quiaochu's sketch.2011-07-14