Use Fourier series for computing $\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}$

Solution 1:

Or you might want to think that

$$\eqalign{ & \omega = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + \cdots = \frac{{{\pi ^2}}}{6} \cr & \frac{\omega }{4} = \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + \frac{1}{{{6^2}}} + \frac{1}{{{8^2}}} + \cdots = \frac{{{\pi ^2}}}{{24}} \cr & \omega - \frac{\omega }{4} = 1 + \frac{1}{{{3^2}}} + \frac{1}{{{5^2}}} + \frac{1}{{{7^2}}} + \cdots = \frac{{{\pi ^2}}}{6} - \frac{{{\pi ^2}}}{{24}} = \frac{{{\pi ^2}}}{8} \cr} $$

Solution 2:

Since the function is odd, we have $\widehat f(2n)=0$ for all integer $n$ and $$\widehat f(2n-1)=\frac 1{2\pi}\frac{\pi}4\left(-\int_{-\pi}^0e^{-i(2n-1)x}dx+\int_0^{\pi}e^{-i(2n-1)x}dx\right)\\\ =\frac 18\left(\frac 1{(2n-1)i}(1-(-1)^{2n-1})+\frac 1{(2n-1)i}(1-(-1)^{2n-1})\right) =\frac 1{2(2n-1)i}.$$ We have $\frac1{2\pi}\int_{-\pi}^{\pi}|f(x)|^2dx=\frac{\pi^2}{16}$ and $|\widehat f(2n-1)|^2=\frac1{4(2n-1)^2}$ so by Parseval equality $$\frac{\pi^2}{16}=\sum_{n\in\mathbb Z}|\widehat f(2n-1)|^2=\sum_{n\in\mathbb Z}\frac1{4(2n-1)^2}=\frac 14 \sum_{n\in\mathbb Z}\frac 1{(2n-1)^2}\\\ =\frac 14\sum_{n\geq 1}\frac 1{(2n-1)^2}+\frac 14\sum_{n\geq 0}\frac 1{(2n+1)^2} =\frac 12\sum_{n\geq 1}\frac 1{(2n-1)^2}. $$ and finally $$\sum_{n=1}^{+\infty}\frac 1{(2n-1)^2}=\frac{\pi^2}8.$$

Solution 3:

You want your computation of the coefficients to be in a convenient form; check that $$|\hat{f}(n)|^2=\begin{cases} (2n)^{-2} & n \text{ odd} \\ \\ 0 & n \text{ even} \end{cases} $$

The major issue I see in the work you posted is that you have an erroneous version of Parseval's:

$$ \hskip 0.3in \sum_{-\infty}^{+\infty} |\hat{f}(n)|= \int_{-\pi}^{\pi} |f(x)|dx \hskip 0.3in \color{Red}{\text{Incorrect}} $$

The correct version is with squares, which makes the computations meaningful:

$$\hskip 0.3in \sum_{-\infty}^{+\infty} |\hat{f}(n)|^2= \int_{-\pi}^{\pi} |f(x)|^2dx \hskip 0.3in \color{LimeGreen}{\text{Correct}} $$

On the left side we'll be adding $(2n)^{-2}$ over the odds twice, and the right side is $\pi^2/16$.

Solution 4:

The Fourier series of $f$ is given by $$ \mathcal{F}f(x)= \sum_{k=1}^{\infty} \frac{1}{2k-1}\sin((2k-1)x), $$ which converges pointwise to $f$ except at $x=0$ (where it is zero). So on $(-\pi, 0)$, we can write $\mathcal{F}f =f$, and integrating both sides of the first equation, we get $$ -\frac{\pi^{2}}{4} = \int_{-\pi}^{0} f(x)\, \mathrm{d}x = \sum_{k=1}^{\infty} \frac{1}{2k-1}\int_{-\pi}^{0}\sin((2k-1)x) = -2\sum_{k=1}^{\infty} \frac{1}{(2k-1)^{2}} $$ which gives the result $$ \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}} = \frac{\pi^{2}}{8} $$