Tricky Sum 2.0

We will start by finding a nice way to express $\sin^2(\omega n \pi)$ . Put $a = \frac 12$ and $b = \frac {\sqrt 3}2$ . We can write $\sin(\omega n \pi) = \frac{e^{i\omega n \pi} - e^{-i\omega n \pi}}{2i} = \frac {e^{-bn\pi} e^{ain\pi} - e^{bn\pi} e^{-ain\pi}}{2i}.$ Notice that $e^{ain\pi} = i^{-n}$ , so $\sin(\omega n \pi) = \frac{e^{-bn\pi} i^{-n} - e^{bn\pi}i^n}{2i}.$ When $n$ is odd, we obtain $\sin(\omega n \pi) = \pm\frac{i(e^{-bn\pi} + e^{bn\pi})}{2i} = \pm \cosh(bn\pi) \implies \sin^2(\omega n \pi) = \cosh^2(bn\pi),$ and when $n$ is even, we obtain $\sin(\omega n \pi) = \pm \frac {(e^{-bn\pi} - e^{bn\pi})}{2i} = \pm i \sinh(bn\pi) \implies \sin^2(\omega n \pi) = -\sinh^2(bn\pi).$ These expressions for $\sin^2(\omega n \pi)$ show the series converges to a real value. While the phrasing of the question presupposes this fact, it is not obvious and part of our solution relies on it, so we state this conclusion here for completeness.

To find the exact value of the series in question, we will use a very specialized form of an identity related to Euler's and Ramanujan's formulas for integer values of the Riemann zeta function. The general identity is given as Theorem 4.2 in this paper , but we will concern ourselves only with the case $m = -2$ : for any $z \in \mathbb{C}$ with $\Im(z) > 0$ , $2z^{-2} \sum_{n = 1}^\infty \frac n{e^{-2ni \pi V(z)} - 1} = 2 \sum_{n = 1}^\infty \frac n{e^{-2ni \pi z} - 1} - \frac 1{12} (1 - z^{-2}) - (2i\pi z)^{-1}, \tag{1}$ where $V$ is the elliptical transformation given by $V(z) = -\frac {(z + 1)}z.$ When we differentiate both sides of (1), we obtain $-4z^{-3} \sum_{n = 1}^\infty \frac n{e^{-2ni\pi V(z)} - 1} + 4i\pi z^{-4} \sum_{n = 1}^\infty \frac {n^2 e^{-2ni\pi V(z)}}{(e^{-2ni\pi V(z)} - 1)^2} = 4i\pi \sum_{n = 1}^\infty \frac {n^2 e^{-2ni\pi z}}{(e^{-2ni\pi z} - 1)^2} - \frac 16 z^{-3} + \frac 1{2i\pi} z^{-2},$ which is equivalent to $-4z^{-3} \sum_{n = 1}^\infty \frac n{e^{-2ni\pi V(z)} - 1} - i\pi z^{-4} \sum_{n = 1}^\infty \frac {-4n^2}{(e^{-ni\pi V(z)} - e^{ni\pi V(z)})^2} = -i\pi \sum_{n = 1}^\infty \frac {-4n^2 }{(e^{-ni\pi z} - e^{ni\pi z})^2} - \frac 16 z^{-3} + \frac 1{2i\pi} z^{-2}. \tag{2}$ Notice the following facts: $\omega^3 = 1, \ \ \ \ \ \ V(\omega) = \omega, \ \ \ \ \ \ e^{-2ni\pi\omega} = (-1)^n e^{bn\pi}, \ \ \ \ \ \ \frac {-4n^2}{(e^{-ni\pi\omega} - e^{ni\pi\omega})^2} = \frac {n^2}{\sin^2(\omega n \pi)}.$ Using these facts, we can evaluate (2) at $z = \omega$ . After a little rearrangement, we obtain $i\pi(1 - \omega^2) \sum_{n = 1}^\infty \frac {n^2}{\sin^2(\omega n\pi)} = 4 \sum_{n = 1}^\infty \frac n{(-1)^n e^{bn\pi} - 1} + \frac 1{2i\pi} \omega - \frac 16.$ Consider the imaginary parts of each side of this equation, keeping in mind that we showed the series on the left to be real. On the left, the imaginary part is $i\pi \sum_{n = 1}^\infty \frac {n^2}{\sin^2(\omega n \pi)} \Re(1 - \omega^2) = \frac 32 i\pi \sum_{n = 1}^\infty \frac {n^2}{\sin^2(\omega n \pi)},$ and on the right, the imaginary part is $\frac 1 {2i\pi} \Re(\omega) = \frac i{4\pi}.$ Equating the imaginary parts shows us the value of the series in question: $\sum_{n = 1}^\infty \frac {n^2}{\sin^2(\omega \pi n)} = \frac i{4\pi} \cdot \frac 2{3i\pi} = \frac 1{6\pi^2}.$ Therefore, the desired value is $1 + 6 - 2 = \boxed{5}$ .