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If we have a circle of radius $r$ with an $n$-gon inscribed within this circle (i.e. with the same circumradius), we can find the difference of the areas using:

$A_n =\overbrace{\pi r^2}^\text{Area of circle}-\overbrace{\frac{1}{2} r^2 n \sin (\frac{2 \pi}{n})}^\text{Area of n-gon} =r^2(\pi-\frac{1}{2} n \sin (\frac{2 \pi}{n}))$

I want to find the following sum (starting with $n=3$, i.e. the $n$-gon is a triangle):

$\Lambda=\sum_{n=3}^{\infty}A_n = r^2\sum_{n=3}^{\infty}(\pi -\frac{1}{2} n \sin (\frac{2 \pi}{n})) = r^2 \lim_{k \rightarrow \infty} (\pi (k-3)-\frac{1}{2} \sum_{n=3}^{k} n \sin (\frac{2 \pi}{n})) $

I have not tested this sum for convergence, but a quick numerical estimate reveals that, if $k=100000, r=1, \Lambda \approx 7.417$. Increasing $k$ gives the same approximation, suggesting convergence.

Does this series converge, and if so, does it have a closed form? Can we find if $\Lambda$ is rational?

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    Perhaps one could express $\sum_{n=3}^k(2\pi-n\sin(2\pi/n))$ as a Riemann sum for some definite integral, and then see about evaluating the integral.2012-04-27

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For convergence, Taylor expand the $\sin$ term; you get

$ \sum_{n=3}^\infty \pi - \frac{1}{2}n\left(\frac{2\pi}{n} - \frac{(\frac{2\pi}{n})^3}{3!} + \ldots \right) $ which is $ \frac{1}{2}\sum_{n=3}^{\infty}\sum_{m = 1}^\infty (-1)^{m+1}\frac{(2\pi)^{2m+1}}{(2m+1)!n^{2m}} $ To prove the outer sum converges, note that each inner sum is alternating and so its first ($m = 1$) term is an overestimate for its value; therefore the sum converges by comparison with $1/n^2$.

I'll bet you can learn more about the value by switching the order of these summations. I tried, but I ended up in a morass of wrongness that I wouldn't have wanted to type up even if it had been right.

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    No problem. I will try and see if I can simplify/solve this sum.2012-04-26