Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$

Question

Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$

a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$

$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$

$b_n = \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$

$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$

The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?

Answer 1

You can prove $$\sum \frac{1}{n^2}=\frac{\pi^2}{6}$$ using Fourier series. Hence, $$\sum\frac{1}{(2n)^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$ Therefore, $$\frac14 \sum \frac{1}{n^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$ This shows $$\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$$

Answer 2

First of all your $b_k$ are wrong, they should be:

$$b_k= \frac{(-1)^{k+1}}{2k}$$

Not that it matters beacause of the following. Second of all notice that $f(x)$ is continuos at zero, which is to say $f(0^+)= f(0^-)=0$. Once you get the expansion right is not that hard, just make $x=0$, easy peasy.

Answer 3

Just a bit late, but I want to post this…

\begin{align} &-\ln(1-x) = \sum_{n=1}^{\infty} \dfrac{x^n}{n} \text{ by Taylor Series.} \\ &x \to e^{ix}: \\ &-\text{Log}(1-e^{ix}) = \sum_{n=1}^{\infty} \dfrac{e^{ixn}}{n}. \\ &-\text{Log}(1-\cos x - i \sin x) = \sum_{n=1}^{\infty} \dfrac {\cos(nx) + i\sin(nx)}{n} \\ \therefore \; & \Im\left(-\text{Log}(1-\cos x - i \sin x)\right) = \sum_{n=1}^{\infty} \dfrac {\sin(nx)}{n} \\ & = -\arg(1-\cos x - i \sin x) = -\arctan\left( \dfrac {-\sin x}{1-\cos x} \right) \\ & = \dfrac {\pi}{2} - \dfrac x 2. \\ \ \\ & \int_0^{\pi} \left( \dfrac {\pi}{2} - \dfrac x 2 \right)dx = \int_0^{\pi} \left( \sum_{n=1}^{\infty} \dfrac {\sin (nx)}{n} \right)dx. \\ & \dfrac {\pi^2}{4} = \sum_{n=1}^{\infty} \dfrac 1 n \int_0^{\pi} \sin(nx)dx \\ &= \sum_{n=1}^{\infty} - \dfrac 1 {n^2} \left( \cos (n\pi) - 1 \right) \\ &= \sum_{n=1}^{\infty} \dfrac 1 {n^2} \left( 1 - \cos (n\pi) - i \sin(n\pi) \right) \\ &= \sum_{n=1}^{\infty} \dfrac 1 {n^2} \left( 1 - e^{in\pi} \right) \\ &= \sum_{n=1}^{\infty} \dfrac 1 {n^2} \left( 1 - (-1)^n \right) \\ &= 2 \cdot \left( \dfrac 1 {1^2} + \dfrac 1 {3^2} + \dfrac 1 {5^2} + \cdots \right). \\ \ \\ \therefore \; & \sum_{n=1}^{\infty} \dfrac 1 {(2n-1)^2} = \dfrac {\pi^2} 8. \blacksquare \end{align}

Answer 4

Substituting $z = \frac{1}{2}$ into this identity

$$ \frac{\pi^2}{\sin^2(\pi z)} = \sum_{m=-\infty}^{\infty} \frac{1}{(z - m)^2} $$

gives

$$ \pi^2 = \sum_{m=-\infty}^{\infty} \frac{1}{\left(\frac{1}{2} - m\right)^2} = 4 \sum_{m=-\infty}^{\infty} \frac{1}{(2m - 1)^2} $$

Therefore

$$ \sum_{m=-\infty}^{\infty} \frac{1}{(2m - 1)^2} = \frac{\pi^2}{4} \tag{1} $$

Separating positive and negative terms:

$$ \sum_{m=-\infty}^{\infty} \frac{1}{(2m - 1)^2} = 2 \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} \tag{2} $$

Equating $(1)$ and $(2)$

$$ 2 \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \frac{\pi^2}{4} $$

Thus,

$$ \boxed{ \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{8} } $$

References

How to prove that $\frac {\pi^2}{ \sin(\pi z)^2 } = \sum_{n=-\infty , n\neq 0 }^{n=\infty} 1/ (z-n)^2 $

Answer 5

$$ \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \sum_{n=1}^{\infty} \int_{0}^{1} \int_{0}^{1} x^{2n-2} y^{2n-2} \, dx \, dy = \int_{0}^{1} \int_{0}^{1} \frac{1}{1 - x^2 y^2} \, dx \, dy = \frac{\pi^2}{8} $$

References

Show that $\int_{0}^{1} \int_{0}^{1} \frac{1}{1 - x^2 y^2} \, dx \, dy = \frac{\pi^2}{8}$

Answer 6

$$ \small \sum_{k=1}^{\infty} \frac{1}{k^2} = \sum_{m=0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{\left(2^m (2n - 1)\right)^2} = \sum_{m=0}^{\infty} \frac{1}{4^m} \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \frac{4}{3} \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \frac{\pi^2}{6} $$ $$ \small \implies \sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = \frac{\pi^2}{8} $$

Answer 7

This would be one way of going about it but using a different fourier series: $$f(x)\ = 1+\sum_{n=1}^{\infty}\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}\cos\left(\frac{\pi nx}{2}\right)$$ Notice how $$\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}$$ is 0 when n is even, so we can just use odd numbers instead with $(2n+1)$. So we can make a new formula with only odds and we can do this by letting $n =(2n+1)$, and because we don't want the even results we can resolve the numerator to be equal to $2$. $$f(x)\ =\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\cos\left(n\pi x\ +\frac{\pi x}{2}\right)$$ Then setting $x = 0$ gives: $$0=\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$ After rearranging: $$\frac{\pi^{2}}{8}=\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$