Just Sum Fourier

$\begin{aligned} \sum_{n=0}^\infty \frac 1{(2n+1)^2} & = \frac 1{1^2} + \frac 1{3^2} + \frac 1{5^2} + \frac 1{7^2} + \frac 1{9^2} + \frac 1{11^2} +\cdots \\ & = \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} \cdots \right) - \left(\frac 1{2^2} + \frac 1{4^2} + \frac 1{6^2} + \frac 1{8^2} + \cdots \right) \\ & = \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} \cdots \right) - \frac 14 \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} \cdots \right) \\ & = \frac 34 \color{#3D99F6} \left(\frac 1{1^2} + \frac 1{2^2} + \frac 1{3^2} + \frac 1{4^2} \cdots \right) & \small \color{#3D99F6} \text{Since Riemann zeta function } \zeta(s) = \sum_{n=1}^\infty \frac 1{n^s} \\ & = \frac 34 {\color{#3D99F6} \zeta (2)} = \frac 34 \times {\color{#3D99F6} \frac {\pi^2}6} = \frac {\pi^2}8 & \small \color{#3D99F6} \text{and } \zeta(2) = \frac {\pi^2}6 \text{ (see reference)} \end{aligned}$

Therefore, $ab = 2\times 8 = \boxed{16}$ .

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Reference: Riemann zeta function

$\begin{aligned}\zeta(x)&={\large\sum}_{n=1}^{\infty}\hspace{2mm}\dfrac{1}{n^x}\\\\ \zeta(2)&={\large\sum}_{n=1}^{\infty}\hspace{2mm}\dfrac{1}{n^2}=\dfrac{{\pi}^2}{6}\\ \dfrac{1}{4}\zeta(2)&={\large\sum}_{n=1}^{\infty}\hspace{2mm}\dfrac{1}{(2n)^2}\\ \zeta(2)-\dfrac{1}{4}\zeta(2)&={\large\sum}_{n=1}^{\infty}\hspace{2mm}\dfrac{1}{n^2}-\dfrac{1}{(2n)^2}\hspace{4mm}\color{#3D99F6}\small\text{all terms with even denominators will be cancelled out}\\ \dfrac{3}{4}\zeta(2)&={\large\sum}_{n=0}^{\infty}\hspace{2mm}\dfrac{1}{(2n+1)^2}\\ \implies{\large\sum}_{n=0}^{\infty}\hspace{2mm}\dfrac{1}{(2n+1)^2}&=\dfrac{3\pi^2}{4\cdot6}\\ &=\dfrac{\pi^2}{8}\\ a&=2,b=8\\ ab&=2\times8=\color{#EC7300}\boxed{\color{#333333}16}\end{aligned}$

$a_{n}=\frac{1}{\pi}\displaystyle \int_{-\pi}^{\pi} |x|Cos(nx) dx$ . The integrand is even so we get. $a_{n}=\frac{2}{\pi}\displaystyle \int_{0}^{\pi} xCos(nx) dx = \frac{2((-1)^n-1)}{n^2\pi} \implies x_{0}=\pi$ .

$|x|= \frac{\pi}{2} + \frac{2}{\pi} \displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n-1}{n^2}Cos(nx)$ on $(-\pi,\pi)$ and the periodic extension of $|x|$ is continuous at $x=\pi$

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

For even $n$ , $\frac{1-(-1)^n}{n^2}=0$ so we need only consider odd $n$ .

$\frac{\pi^2}{4} = \displaystyle \sum_{n=0}^{\infty} \frac{2}{(2n+1)^2}$

$\implies \displaystyle \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8} \implies a=2, b=8$

Finally $a\times b = 2\times 8 = \boxed{16}$ .

It's actually ((π^2)/8)