A tricky sum to infinity
ETA: These bounds are wrong, as $\frac{kx}{(k^2+x)^2}$ is not monotone in $k$. For a fixed version of this answer, see robjohn's answer here.
Notice that, for fixed $x$, your sum is less than $$\int_0^\infty \frac{kx}{(k^2+x)^2} \, dk=\frac{1}{2}\, ,$$ and greater than $$\int_1^\infty \frac{kx}{(k^2+x)^2} \, dk=\frac{x}{2(1+x)} \, ,$$ and then apply the squeeze theorem.