I can show that $\sum^\infty_{k=1}{{1}\over{k^2}} = {{\pi^2}\over{6}}$ without to much hassle just using two representations of ${{sin(x)}\over{x}}$ but I cant find anything nearly as simple when I try to determine what $\sum^\infty_{k=1}{{1}\over{k^2+1}}$ equals.

Does there exist an elementary proof? or must I resort to using integrals?