The series is
$$\tan^{-1} (\frac{1}{(2)(1^2)}) + \tan^{-1} (\frac{1}{(2)(2^2)})+ \tan^{-1} (\frac{1}{(2)(3^2)}).....$$
I know the series will telescope, but I don’t know how it would work. These problems usually reduce to $\tan^{-1} a- \tan^{-1} b$, but that’s not happening here
You can prove by induction the $n$th partial sum is $\arctan\frac{n-1}{n}$, so the series has limit $\arctan1=\frac{\pi}{4}$. So your telescoping strategy works viz. $a=\frac{k}{k+1},\,b=\frac{k-1}{k}$.