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Sum over all positive rationals

For a rational number x that equals \frac{a}{b} in lowest terms , let f(x)=ab. Prove that:

    \[\sum_{x \in \mathbb{Q}^+} \frac{1}{f^2(x)} = \frac{5}{2}\]


First of all we note that

    \[F(s) = \sum_{x \in \mathbb{Q}^+} \frac{1}{f^s(x)} = \sum_{\substack{a,b=1 \\ \gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s}\]

Moreover for s>1 we have that

    \begin{align*} \zeta^2(s) &= \left ( \sum_{a=1}^{\infty} \frac{1}{a^s} \right )^2 \\ &=\sum_{a, b=1}^{\infty} \frac{1}{(ab)^s} \\ &=\sum_{d=1}^{\infty} \sum_{\substack{a, b=1 \\\gcd(a, b)=d}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^{2s}} \sum_{\substack{a, b=1 \\\gcd(a, b)=1}}^{\infty} \frac{1}{\left ( ab \right )^s} \\ &= \zeta(2s) F(s) \end{align*}

Hence for s=2 we have that

    \[F(2) = \frac{5}{2}\]

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