Add 'Til You Reach $\infty$

The key integral is $\int_0^\infty \frac{dx}{x^n + x^{1/n}} = \frac{n\pi}{n^2-1} \csc\left( \frac{\pi}{n+1} \right).$ This can be proved by contour integration--I edited my answer to give the details below.

Then the sum is $\begin{aligned} \sum_{n=2}^\infty \frac1{n\pi} \sin\left( \frac{\pi}{n+1} \right) \frac{n\pi}{n^2-1} \csc\left( \frac{\pi}{n+1} \right) &= \sum_{n=2}^\infty \frac1{n^2-1} \\ &= \sum_{n=2}^\infty \left( \frac{1/2}{n-1} - \frac{1/2}{n+1} \right) \\ &= \left( \frac12 - \frac16 \right) + \left( \frac14 - \frac18 \right) + \left( \frac16 - \frac1{10} \right) + \cdots \\ &= \frac12 + \frac14 = \frac34 \end{aligned}$ (the series telescopes ), so the answer is $\fbox{12}.$

Edit: here is the derivation of the integral formula. First make the substitution $u = z^{1/n}$ to change the integral to $\int_0^\infty \frac{n u^{n-2} \, du}{u^{n^2-1} + 1}.$

Let $\zeta = \text{exp}(\pi i/(n^2-1)),$ so $\zeta$ is a primitive $2(n^2-1)$ th root of unity. The poles of $\frac{n u^{n-2}}{u^{n^2-1} + 1}$ are the odd powers of $\zeta.$ Consider the pie-slice-shaped contour cut out by the two rays $R_1, R_2$ given by $y=0, 0 \le x \le N$ and $y = \zeta^2 x, 0 \le x \le M$ (where $M$ is chosen so the ray has length $N$ ) and the circular arc connecting them. The integral over the arc goes to zero as $N$ gets large, so we have, by the residue theorem, $\begin{aligned} 2\pi i \text{Res}_{u=\zeta} \frac{nu^{n-2}}{u^{n^2-1}+1} &= \int_0^\infty \frac{n u^{n-2} \, du}{u^{n^2-1} + 1} - \int_0^{\infty} \frac{n(\zeta^2 v)^{n-2} \zeta^2 \, dv}{v^{n^2-1}+1} \\ 2\pi i \frac{n\zeta^{n-2}}{(n^2-1)\zeta^{n^2-2}} &= I - \zeta^{2n-2} I \end{aligned}$ where $I$ is the integral we want. So $I = \frac{2\pi i n}{n^2-1} \frac{\zeta^{n-n^2}}{1-\zeta^{2n-2}} = \frac{2\pi i n}{n^2-1} \frac{\zeta^{n-1}\zeta^{1-n^2}}{1-\zeta^{2n-2}} = \frac{2\pi i n}{n^2-1} \frac{\zeta^{n-1}}{\zeta^{2n-2}-1}$ because $\zeta^{1-n^2} = -1.$ This gives $I = \frac{2\pi i n}{n^2-1} \frac1{\zeta^{n-1} - \zeta^{-(n-1)}} = \frac{2\pi i n}{n^2-1} \frac1{2i \sin\left(\frac{2\pi(n-1)}{2(n^2-1)}\right)} = \frac{\pi n}{n^2-1} \csc\left( \frac{\pi}{n+1} \right)$ as desired.