Evaluating the sum $\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n^{2}}$

Solution 1:

On the contour $C_N$, we have $\lvert z^2\rvert \geqslant \left(N+\frac12\right)^2\pi^2$, so if we can bound the modulus of the sine factor below, say $\lvert \sin z \rvert \geqslant \varepsilon > 0$ on $C_N$, the standard estimate yields

$$\left\lvert \int_{C_N} \frac{dz}{z^2\sin z}\right\rvert \leqslant \frac{4\cdot(2N+1)\pi}{\left(N+\frac12\right)^2\pi^2\varepsilon} = \frac{8}{\left(N+\frac12\right)\pi\varepsilon} \xrightarrow{N\to\infty} 0.$$

But with $z = x+iy$, we have $\sin z = \sin x \cos (iy) + \cos x\sin (iy) = \sin x \cosh y + i\cos x\sinh y$, so

$$\lvert \sin z\rvert^2 = \sin^2 x \cosh^2 y + \cos^2 x \sinh^2y = \sin^2 x + \sinh^2 y.$$

Now, on the vertical sides of the contour $C_N$ ($x = \pm \left(N+\frac12\right)\pi$) we have $\sin x = \pm 1$, so $\lvert \sin z\rvert \geqslant 1$ on those. On the horizontal sides, we have $\lvert \sin z\rvert \geqslant \sinh \left(\left(N+\frac12\right)\pi\right) \geqslant \sinh \frac{\pi}{2} > 2$, so we can choose $\varepsilon = 1$ in the above.