Astonishing result

Calculus Level 4

i=1j=1k=11(ijk)2=πab, \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \dfrac{1}{(ijk)^2} = \dfrac{\pi^a}{b},

where aa and bb are integers. Find ba.\frac{b}{a}.

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