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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 $\sum_{n=1}^\infty {1\over n^2}={\pi\over 6}$?

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up vote 5 down vote accepted

From your first result, $$\sum_{k=1}^\infty \frac{1}{n^2+k^2} = \frac{1}{2} \left(\frac{\pi}{k \tanh(\pi k)} - \frac{1}{k^2} \right)$$ Now use $$\frac{1}{\tanh(\pi i)} = \frac{\cosh(\pi k)}{\sinh(\pi k)} = \frac{1 + \pi^2 k^2/2 + \ldots}{\pi k + \pi^3 k^3/6 + \ldots}$$

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