user 85795

$f$ is defined as $$ f(x)=\sum_{k=1}^\infty\frac{\arctan(kx)}{k^2}\tag1 $$ For $0\lt x\le1$, the Mean Value Theorem says that there is a $h\in(0,x)$ so that $$ \begin{align} \frac{f(x)-f(0)}x &=f'(h)\tag{2a}\\ &=\sum_{k=1}^\infty\frac1{k\left(1+k^2h^2\right)}\tag{2b}\\ &\ge\sum_{k=1}^{\lfloor1/h\...