The area of a regular n-gon with edge length $s$ is given by $$\frac{ns^2}{4\tan\left(\frac{\pi}{n}\right)}$$
Since these are areas I presumed that they had some sort of squared growth rate. So in the same spirit as $\zeta(2)$ I wondered what the sum of their reciprocals is.
1) Thus, can we find a closed form in terms of $s$ for the series $$\sum_{n=3}^\infty\frac{4\tan\left(\frac{\pi}{n}\right)}{ns^2}$$
A similar question concerns the reciprocal of the areas of consecutive regular polygons without anything fixed. Rather, as the number of sides increases, the radius increases at the same rate.
The area of a regular n-gon in terms of the circumradius, $R$ is given by $$\frac12nR^2\sin\left(\frac{2\pi}{n}\right)$$
2) Thus can we find a closed form for the series $$\sum_{n=3}^\infty\frac{2}{n\left(n+k\right)^2\sin\left(\frac{2\pi}{n}\right)}$$ For integer $k\ge -2$. Either a specific choice for $k$ or general $k$ would be acceptable for answer.
Alongside this question, we might try experimenting with other radius growth rates that cause the series to converge.
1) In the same spirit of Basel problem, the logarithmic derivative of the Weierstrass product for the cosine function gives us the identity $$z\tan(z)=2\sum_{m\geq 1}\frac{(4^m-1)\,\zeta(2m)}{\pi^{2m}}\,z^{2m} \tag{A}$$ for any $z$ such that $|z|<\frac{\pi}{2}$, hence by replacing $z$ with $\frac{\pi}{n}$ and summing over $n\geq 3$ we get:
$$\begin{eqnarray*}\sum_{n\geq 3}\frac{1}{n}\,\tan\left(\frac{\pi}{n}\right)&=&\frac{2}{\pi}\sum_{m\geq 1}(4^m-1)\,\zeta(2m)\left(\zeta(2m)-1-\frac{1}{4^m}\right)\\&=&\frac{2}{\pi}\sum_{m\geq 1}(4^m-1)\zeta(2m)\int_{0}^{+\infty}\frac{x^{2m-1}/(2m-1)!}{e^{2x}(e^x-1)}\,dx\\&=&\frac{2}{\pi}\iint_{(0,+\infty)^2}\frac{I_0(2\sqrt{2xy})-J_0(2\sqrt{2xy})}{e^{2x}(e^x-1)(e^y-e^{-y})}\,dx\,dy\end{eqnarray*}\tag{B} $$ which is pretty simple to approximate numerically, due to the well-known asymptotic behavior of the Bessel functions $I_0,J_0$ both in a right neighbourhood of the origin and in a left neighbourhood of $+\infty$. On the other hand, I would not bet on a "nice" closed form for the LHS of $(B)$, $\approx 1.56439140145850977221273$.
2) can be managed through similar techniques, and leads to similar issues. The Taylor series of $\frac{z}{\sin z}$ can be computed by differentiating $\log\tan\frac{x}{2}$, where $\tan\frac{x}{2}$ has a well-known Weierstrass product.