how can we show $\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + …$?

Question

Let $f(x) = \frac4\pi \cdot (\sin x + \frac13 \sin (3x) + \frac15 \sin (5x) + \dots)$. If for $x=\frac\pi2$, we have
$$f(x) = \frac{4}{\pi} ( 1 - \frac13 +\frac15 - \frac17 + \dots) = 1$$ then obviously : $$ 1 - \frac13 +\frac15 - \frac17 + \dots=\frac{\pi}{4}$$ Now how can we prove that: $$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$

Answer 1

To prove the sum $\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dotsc$ I don't see how it could be useful to refer to $$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\dotsc=\frac{\pi}{4}$$ It might, however, be convenient to recall the famous sum of Euler: $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dotsc=\frac{\pi^2}{6}$$ and to think of possible variations of this sum in order to obtain your desired sum...

What terms are missing between these two sums?

How is this difference related to one of the original sums?

$\dotsc$

Answer 2

From the Basel Problem, we have $$\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2}\dots$$ $$\frac{\pi^2}{24} = \frac{\pi^2}{6\cdot 2^2} = \frac{1}{2^2} + \frac{1}{4^2} + \frac{1}{6^2}\dots$$

so that

$$\begin{align}\frac{\pi^2}{8} &= \frac{\pi^2}{6} - \frac{\pi^2}{24}\\&=\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \dots \end{align}$$

Answer 3

A possible short proof comes from Parseval's identity.

It is well known that the Fourier sine series of the $\text{sign}$ function over $(-\pi,\pi)$ is given by: $$ \text{sign}(x) = \frac{4}{\pi}\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} \tag{1}$$ hence by Parseval's identity: $$ 2\pi = \int_{-\pi}^{\pi} 1\,dx = \frac{16}{\pi}\sum_{n\geq 0}\frac{1}{(2n+1)^2}\tag{2}$$ and the claim readily follows.

Answer 4

Hint: Prove that $\bigg(1+\dfrac13-\dfrac15-\dfrac17+\dfrac19+\dfrac1{11}\pm\cdots\bigg)^2=1+\dfrac1{3^2}+\dfrac1{5^2}+\dfrac1{7^2}+\dfrac1{9^2}+\cdots$

Answer 5

This is a well known double integral proof by Beukers, Kolk, and Calabi. First consider the double integral:

$$\int_{0}^{1}\int_{0}^{1} \frac{1}{1-x^2y^2} dydx.$$

Since $0<x,y<1$, rewrite the integrand as a geometric series:

$$\frac{1}{1-x^2y^2}=\sum_{n=0}^{\infty}(xy)^{2n}.$$

Now notice: $$\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^{\infty}(xy)^{2n}dydx$$

is the same as:

$$\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}(xy)^{2n}dydx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}.$$

Make the change of variables: $$x=\frac{\sin(u)}{\cos(v)},y=\frac{\sin(v)}{\cos(u)}.$$

The Jacobian Determinant is:

$$\det\frac{\partial (x,y)}{\partial(u,v)}=\begin {vmatrix} \frac{\cos(u)}{\cos(v)}&&\frac{\sin(u)\sin(v)}{(\cos(v))^2} \\ \frac{\sin(v)\sin(u)}{(\cos(u))^2}&&\frac{\cos(v)}{\cos(u)}\\\end{vmatrix}=1-x^2y^2,$$

which cancels with the integrand, and the region of integration is the open iscosceles triangle formed by the inequalities: $$0<u+v<\frac{\pi}{2},0<u,v<\frac{\pi}{2}.$$

Using either geometry or evaluating the double integral:

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}-v}1dudv,$$

the area of the isosceles triangle is $\frac{\pi^2}{8}$. So the result is that:

$$\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$$