Sum over all positive rationals

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

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

Solution

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

$F(2) = \frac{5}{2}$

Sum over all positive rationals

They might interest you:

An indefinite integral

An integral

Floor of square root

Limit of a multiple integral

Divergent Möbius series

A contour integral

Arctan integral

Not Lebesgue integrable function

The line is the shortest path between two points

Leave a Reply Cancel reply