# Convergence and closed form of this infinite series?

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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Expanding the sine as a power series gives $$A_n = \frac{2 \pi ^3 r^2}{3 n^2} - \frac{2 \pi ^5 r^2}{15 n^4} + \frac{4 \pi ^7 r^2}{315 n^6} - \frac{2 \pi ^9 r^2}{2835 n^8} + \cdots,$$ from which we can see that $$\lim_{n \to \infty} \frac{A_n}{2 \pi^3 r^2/(3 n^2)} = 1.$$ Since $\sum \frac{1}{n^2}$ converges, $\sum A_n$ converges by the limit comparison test. –  Antonio Vargas Apr 25 '12 at 20:48
Any ideas about finding the closed form of this series? Is it even possible? –  Argon Apr 26 '12 at 1:13
I am getting different numerical values (6.something) when approximating your series. –  user29743 Apr 26 '12 at 16:25
@countinghaus So does WolframAlpha (wolframalpha.com/input/?i=sum_k%3D3%5Einfinity+pi-.5*k*sin%282*pi%2F‌​k%29). However, on that same page, when you click "more terms" on the graph of "Partial sums" several times, it is clear the sum is greater then 7. My approximation is from a quick C++ program I coded. –  Argon Apr 26 '12 at 20:07
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. –  Gerry Myerson Apr 27 '12 at 0:21

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 $1/n^2$.
You can, I just gave up because I couldn't make it much nicer. There were mistakes in the earlier version I had up here by the way, sorry about that. Also, I got rid of $r$ since we will just multiply the final answer by $r^2$. –  user29743 Apr 26 '12 at 5:45