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sum from n = 1 to infinity of 1/n^(2p )

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Solution Summary

The solution explains how one compute the zeta function of even arguments, i.e. summations of the form sum from n = 1 to infinity of 1/n^(2p ) for integer p, using the identity:

pi^2/sin^2(pi z) = sum over n from minus to plus infinity of 1/(z-n)^2.

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We have the identity:

pi^2/sin^2(pi z) = sum over n from minus to plus infinity of 1/(z-n)^2

and we want to extract the summations of the form:

sum from n = 1 to infinity of 1/n^p

for even p from this.

We can do this by considering the identity in the neighborhood of z = 0. If we put z = 0, then we get almost the desired summation for p = 2, however, we obviously do not want to n = 0 term in the summation in that case. We also see that the left hand side is undefined for z = 0, the function pi^2/sin^2(pi z) has a pole there. We can deal with this problem by subtracting from both sides of the identity, the singular part of the Laurent expansion around z = 0. Then the limit for z to zero exists and what we are subtracting from the summation can be interpreted as removing the n = 0 term in the z to zero limit. In case of p = 4, 6 etc. we can differentiate both sides of the identity, and subtract singular terms of the Laurent expansion from both sides. Now, while this will certainly work, it is still very tedious to proceed in this way. So, we need to think of doing this computation efficiently.

Clearly, what we would be doing if we were to go about computing the summations as described above, is extracting the constant term of the Laurent expansion. So, we don't actually need to subtract the singular terms and then take the limit of z to zero ...

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