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

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: $$\frac{\pi^2}{8} = 1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$$

• Please show your effort. Also, fix the title. The question is not if this is a calculus question or not, I presume. – mickep Sep 28 '15 at 12:52
• Correct me if I'm confused, but how can you tell that $f\left(\frac{\pi}{2}\right) = 1$? – Yiyuan Lee Sep 28 '15 at 13:36
• @user160867 is this the question given in FIITJEE package?I have seen it there – Snehil Sinha Sep 28 '15 at 13:41
• That is the part that is not clear for me too. I wanted to suppose it as an assumption and obtain $1 + \frac1{3^2} +\frac1{5^2} + \frac1{7^2} + \dots$ in terms of $1 - \frac1{3} +\frac1{5} - \frac1{7} + \dots$ – user160867 Sep 28 '15 at 13:41
• I do not know about FITJEE package... – user160867 Sep 28 '15 at 13:42

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}

• Thanks for suggestion. Any method with use of f(x) is appreciated too. – user160867 Sep 28 '15 at 13:19

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?

$\dotsc$

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:

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

• That change of variables is epic – Klangen Aug 2 at 21:56

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.

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