Here is another approach :
We can easily prove that $ \left(\forall x\in\left[0,\frac{1}{2}\right]\right),\ \frac{2}{\pi}\arctan{\left(2x\right)}-x\geq 0 $ (By differentiation the function $ x\mapsto\frac{2}{\pi}\arctan{\left(2x\right)}-x $, studying its variations, etc...)
Thus, if $ n $ is a positive integer, we have the following : \begin{aligned}\sum_{k=2}^{n}{\frac{1}{k^{2}}}\leq\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{2}{k^{2}}\right)}}&=\frac{2}{\pi}\sum_{k=2}^{n}{\arctan{\left(\frac{\left(k+1\right)-\left(k-1\right)}{1+\left(k+1\right)\left(k-1\right)}\right)}}\\&=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{\left(k+1\right)}-\arctan{k}\right)}+\frac{2}{\pi}\sum_{k=2}^{n}{\left(\arctan{k}-\arctan{\left(k-1\right)}\right)}\\ &=\frac{2}{\pi}\left(\arctan{\left(n+1\right)}-\arctan{2}\right)+\frac{2}{\pi}\left(\arctan{n}-\frac{\pi}{4}\right)\\ &\leq\frac{3}{2}-\frac{2}{\pi}\arctan{2}<1\end{aligned}
To get to the last line we upper bounded the $ \arctan $s by $ \frac{\pi}{2} \cdot $