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

I am tasked to evaluate the sum $$\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n^{2}}$$ Using contour integration. This is what I've done so far.

Let $C_{N}$ be the square defined by the lines $x=\pm(N+\tfrac{1}{2})\pi$ and $y=\pm(N+\tfrac{1}{2})\pi$. Let $f=\tfrac{1}{z^{2}\sin(z)}$. I was able to prove that $$\int_{C_{N}}\dfrac{1}{z^{2}\sin(z)}dz=2\pi i\left[\dfrac{1}{6}+2\sum_{n=1}^{N}\dfrac{(-1)^{n}}{\pi^{2}n^{2}} \right].$$

By Cauchy's residue theorem. So now if I could prove that he integral converges to zero then I would be done. The problem is that I can't seem to get an upper bound on the integral. I've having trouble with the fact that $|f|$ is unbounded within $C_{N}$ so ML inequality doesn't help me. Any hints would be great. Thanks.

• Was that function given to you or you came up with it? – DonAntonio Dec 28 '13 at 0:37
• Are you sure the index starts at 0? Because the first term is in determinant. – Ali Caglayan Dec 28 '13 at 0:47
• *indeterminate ${}$ – anon Dec 28 '13 at 0:51
• @DonAntonio I was given the function. – TheNumber23 Dec 28 '13 at 1:07

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.

$Res(\frac {1}{z^2sin z},0)=0$. Write it like this $$f(z)=\frac {1}{z^2sin z}=\frac {1}{z^2(z-\frac {z^3}{3}+\frac {z^5}{5}-...)}=\frac {1}{z^3(1-\frac {z^2}{3}+\frac {z^4}{5}-...)}$$. So $f$ has a pole of order 3 at $0$.

• I've already proved the integral result. And calculated the residue at 0. – TheNumber23 Dec 28 '13 at 1:08