2015 September 12
originally posted on facebook

Fun fact: the sum of 1/(ab)^2 over all relatively prime pairs of positive integers a, b is rational. This is not true if you remove the “relatively prime” restriction.

Short proof… that sum equals

\prod_{\text{prime } p} \left(1 + \frac 2{p^2} + \frac 2{p^4} + \frac 2{p^6} + \cdots\right) = \prod \frac {1 + 1 / p^2}{1 - 1 / p^2} = \prod \frac {1 - 1 / p^4}{(1 - 1 / p^2)^2} = \frac {\zeta(2)^...

where \zeta is the Riemann zeta function. Since \zeta(k) is equal to \pi^k times a rational when k is positive even, the \pis cancel and the result is rational, in fact 5/2. If you remove the restriction you get \zeta(2)^2 = \pi^4 / 36 which is irrational. You can make other combinations of \zetas so that the \pis cancel to get other rational infinite summations, for example the sum of 1/(abc)^2 where a is squarefree and divides b should give \zeta(2)^2 \zeta(4) / \zeta(8) = 35 / 12, and in both examples replacing the exponent 2 by any even integer also works.

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