A convergent series

Note that $f'(x)=\sum_{k=1}^{\infty}\frac{2k\sin(kx)\cos(kx)}{k^2}=\sum_{k=1}^{\infty}\frac{\sin(2kx)}{k}=\sum_{k=1}^{\infty}\frac{e^{i2kx}-e^{-i2kx}}{2ik}=\frac{-\ln\left(1-e^{i2x}\right)+\ln\left(1-e^{-i2x}\right)}{2i}=\frac{\ln\left(-e^{-2ix}\right)}{2i}$

Now $f'(x)= 0$ implies that $-e^{-i2x}=1$ or equivalently, $x=\frac{(2n-1)\pi}{2}$ . For $x\in (0,\pi)$ , we have $x=\frac{\pi}{2}$ .

Now $f'(x)=\frac{\ln\left(-e^{-2ix}\right)}{2i}$ implies that $f''(x)=-1<0$ for all $x\in (0,\pi)$ which means that $f$ is concave downwards and $f\left(\frac{\pi}{2}\right)$ is the maximum value.

Finally $f\left(\frac{\pi}{2}\right)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8} \approx 1.2337$