Coprime Collections (8/26 – 9/8)
Find the value of
$$
\lim_{N \to \infty} \sum_{\substack{1 \leq a, b \leq N \\ \gcd(a,b) = 1}} \frac{1}{a^2b^2}
$$
where the summation runs from all values of \(1 \leq a \leq N, 1 \leq b \leq N, \gcd(a,b) = 1\)
given
$$
\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}, \sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}
$$
Solution:
Let the sum we wish to compute equal to \(S\).
Consider the sum
\[\sum_{a=1}^\infty \sum_{b=1}^\infty \frac{1}{a^2b^2}\]
This is equal to \[\sum_{a=1}^\infty \frac{1}{a^2} \sum_{b=1}^\infty \frac{1}{b^2} = \frac{\pi^4}{36}\]
We can rewrite this into
$$\mathop{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}}_{\gcd(a,b) = 1} \frac{1}{a^2b^2} + \mathop{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}}_{\gcd(a,b) = 2} \frac{1}{a^2b^2} + \dots$$
Using the fact that if \(\gcd(a,b) = k, \gcd(\frac{a}{k}, \frac{b}{k}) = 1\), we can conclude that
$$\mathop{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}}_{\gcd(a,b) = k} \frac{1}{a^2b^2} = \mathop{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}}_{\gcd(a,b) = 1} \frac{1}{(ka)^2(kb)^2} = \frac{1}{k^4}\mathop{\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}}_{\gcd(a,b) = 1} \frac{1}{a^2b^2}$$
Thus, we get that
\[\frac{\pi^4}{36} = \sum_{a=1}^\infty \sum_{b=1}^\infty \frac{1}{a^2b^2} = S + \frac{S}{2^4} + \frac{S}{3^4} + \dots = S(1 + \frac{1}{2^4} + \dots) = S \cdot \frac{\pi^4}{90}\]
$$\therefore S = \frac{5}{2}$$