Given that the series $\sum_{-\infty}^\infty {1\over n^2+k^2}={\pi\over k \tanh (\pi k)}$ and $\sum_{-\infty}^\infty {1\over (n+k)^2}={\pi^2\over \sin^2 (\pi k)}$
How might we take the limit of $k\to 0$ so that we can get $\sum_{n=1}^\infty {1\over n^2}={\pi\over 6}$?