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Do problem #3 on page 153, which asks us to compute the alternating sum of the reciprocals of the squares:
$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} = \frac{\pi^2}{12}.$$
The full Fourier series for $f(x)=x^2$ on $[-\pi,\pi]$ is
$$\frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos(nx)+b_n\sin(nx)\right),$$
where
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx$$
and
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\sin(nx)dx.$$
First, we find $\frac{a_0}{2}$, taking advantage of the fact that for an even function $g(x)$,
$$\int_{-L}^{L}g(x)dx = 2\int_{0}^{L}g(x)dx$$:
$$\frac{a_0}{2} = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{1}{\pi}\int_{0}^{\pi}x^2dx = \left[\frac{x^3}{3\pi}\right]_0^{\pi} = \frac{\pi^2}{3}.$$
Now we find $a_n,$ similarly taking advantage of the fact that $x^2cos(nx)$, as the product of two even functions, is even:
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx = \frac{2}{\pi}\int_{0}^{\pi}x^2\cos(nx)dx.$$
We solve this integral by parts, setting $u=x^2$, $du=2xdx$, $v=\frac{\sin(nx)}{n}$, and $dv=\cos(nx)dx$:
$$ \frac{2}{\pi}\int_{0}^{\pi}x^2\cos(nx)dx = \frac{2}{n\pi}\left(\left[x^2\sin(nx)\right]_0^{\pi} - 2\int_{0}^{\pi}x\sin(nx)dx\right) = -\frac{4}{n\pi}\int_0^{\pi}x\sin(nx)dx.$$
Here we integrate by parts again, setting $u=x$, $du=dx$, $v=-\frac{\cos(nx)}{n}$, and $dv=\sin(nx)$:
$$-\frac{4}{n\pi}\int_0^{\pi}x\sin(nx)dx = \frac{4}{\pi n^2}\left(\left[x\cos(nx)\right]_0^{\pi} - \int_0^{\pi}\cos(nx)dx\right) = \frac{4}{\pi n^2}\left(\pi\cos(nx)-0\right) = \frac{4(-1)^n}{n^2}.$$
Therefore,
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx = \frac{4(-1)^n}{n^2}.$$
Finally, since the product of an even function and an odd function is odd, and for an odd function $h(x),$
$$\int_{-L}^Lh(x)dx = 0,$$
we can conclude that
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\sin(nx)dx = 0.$$
So the Fourier series for $x^2$ on $[-\pi,\pi]$ is
$$x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx).$$
Setting $x=0$:
$$0 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2},$$
so
$$\frac{\pi^2}{3} = -4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} = 4\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}.$$
Dividing both sides by 4:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}.$$
Comments
The full Fourier series for $f(x)=x^2$ on $[-\pi,\pi]$ is
$$\frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos(nx)+b_n\sin(nx)\right),$$
where
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx$$
and
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\sin(nx)dx.$$
First, we find $\frac{a_0}{2}$, taking advantage of the fact that for an even function $g(x)$,
$$\int_{-L}^{L}g(x)dx = 2\int_{0}^{L}g(x)dx$$:
$$\frac{a_0}{2} = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx = \frac{1}{\pi}\int_{0}^{\pi}x^2dx = \left[\frac{x^3}{3\pi}\right]_0^{\pi} = \frac{\pi^2}{3}.$$
Now we find $a_n,$ similarly taking advantage of the fact that $x^2cos(nx)$, as the product of two even functions, is even:
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx = \frac{2}{\pi}\int_{0}^{\pi}x^2\cos(nx)dx.$$
We solve this integral by parts, setting $u=x^2$, $du=2xdx$, $v=\frac{\sin(nx)}{n}$, and $dv=\cos(nx)dx$:
$$ \frac{2}{\pi}\int_{0}^{\pi}x^2\cos(nx)dx = \frac{2}{n\pi}\left(\left[x^2\sin(nx)\right]_0^{\pi} - 2\int_{0}^{\pi}x\sin(nx)dx\right) = -\frac{4}{n\pi}\int_0^{\pi}x\sin(nx)dx.$$
Here we integrate by parts again, setting $u=x$, $du=dx$, $v=-\frac{\cos(nx)}{n}$, and $dv=\sin(nx)$:
$$-\frac{4}{n\pi}\int_0^{\pi}x\sin(nx)dx = \frac{4}{\pi n^2}\left(\left[x\cos(nx)\right]_0^{\pi} - \int_0^{\pi}\cos(nx)dx\right) = \frac{4}{\pi n^2}\left(\pi\cos(nx)-0\right) = \frac{4(-1)^n}{n^2}.$$
Therefore,
$$a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos(nx)dx = \frac{4(-1)^n}{n^2}.$$
Finally, since the product of an even function and an odd function is odd, and for an odd function $h(x),$
$$\int_{-L}^Lh(x)dx = 0,$$
we can conclude that
$$b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}x^2\sin(nx)dx = 0.$$
So the Fourier series for $x^2$ on $[-\pi,\pi]$ is
$$x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx).$$
Setting $x=0$:
$$0 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2},$$
so
$$\frac{\pi^2}{3} = -4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} = 4\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}.$$
Dividing both sides by 4:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}.$$