This is not homework.

My question is:

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> Prove or disprove: $$\lim\_{n \to +\infty} \frac{1}{n^2} \sum\_{a,b=1}^n \frac{1}{ \mathrm{gcd} (a,b)} = \frac{\zeta(3)}{\zeta(2)}$$
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This would represent the probability as $n \to +\infty$ that, after having picked randomly $a,b,c \in \{ 1, \dots ,n \}$, the line
$$ax+by+c=0$$
has some integer coordinate point $(x,y \in \Bbb Z^2)$. Indeed, fixed $a,b \in \{ 1, \dots ,n \}$, then $c=-ax-by$ for some $x,y \in \Bbb Z$ is equivalent to $$c \mathrm{\ is \ a\ multiple\ of\ gcd}(a,b)$$
because of Bezout identity; this event happens with probability $1/\gcd (a,b)$.

What I tried first: clearly
$$0 \le \frac{1}{n^2} \sum\_{a,b=1}^n \frac{1}{ \mathrm{gcd} (a,b)} \le \frac{1}{n^2} \sum\_{a,b=1}^n 1 = 1$$
so, our limit belongs to $[0,1]$ (if it exists).

Then I made some euristic argument:

I know that the probability, as $n \to +\infty$, that two random numbers $a,b\in \{ 1, \dots ,n \}$ are coprime is $$\zeta(2)^{-1} = \frac{6}{\pi^2},$$ thus the probability that $\gcd(a,b)=d$ is equal to
$$\frac{\zeta(2)^{-1}}{d^2} = \frac{6}{d^2\pi^2}$$
Thus our probability is (and here is where I have doubts in my argument)
$$\sum\_{d=1}^n \ \frac{\zeta(2)^{-1}}{d^2} \to \frac{1}{\zeta(2)} \cdot \zeta(2) =1$$

EDIT: or maybe my last step should be
$$\sum\_{d=1}^n \ \frac{\zeta(2)^{-1}}{d^2} \frac{1}{d} \to \frac{\zeta(3)}{\zeta(2)}$$