Fdtxjr

This is not homework.

My question is:

Prove or disprove: $$lim_n to +infty frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) = fraczeta(3)zeta(2)$$

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
$$0 le frac1n^2 sum_a,b=1^n frac1 mathrmgcd (a,b) le frac1n^2 sum_a,b=1^n 1 = 1$$
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,bin 1, dots ,n $ are coprime is $$zeta(2)^-1 = frac6pi^2,$$ thus the probability that $gcd(a,b)=d$ is equal to
$$fraczeta(2)^-1d^2 = frac6d^2pi^2$$
Thus our probability is (and here is where I have doubts in my argument)
$$sum_d=1^n fraczeta(2)^-1d^2 to frac1zeta(2) cdot zeta(2) =1$$

EDIT: or maybe my last step should be
$$sum_d=1^n fraczeta(2)^-1d^2 frac1d to fraczeta(3)zeta(2)$$

There is no need to involve probabilities, just a bit of analytic number theory suffices. I don't know how familiar you are with $L$-functions, so I'll leave some details open for now. If any part needs clarification, let me know.

For every positive integer $n$ define
$$f(n):=sum_a,b=1^nfrac1gcd(a,b)
qquadtext and qquad
g(n):=sum_dmid nfracvarphi(tfracnd)d$$ so that for all $n>1$ we have
begineqnarray*
f(n)-f(n-1)
&=&-frac1n+2sum_c=1^nfrac1gcd(c,n)\
&=&-frac1n+2sum_dmid nfracvarphi(tfracnd)d\
&=&-frac1n+2g(n).
endeqnarray*
Because $lim_ntoinftyfrac1n^2sum_k=1^n-frac1k$, this already shows that
$$lim_ntoinftyfrac1n^2f(n)=2lim_ntoinftyfrac1n^2sum_k=1^ng(n).$$
Note that $g$ is a multiplicative function, and in fact $g=iotaastvarphi$ where $iota(n):=tfrac1n$. This immediately shows that the $L$-function associated to $g$ converges for all $sinBbbC$ with $operatornameRe(s)>2$ and that for all such $sinBbbC$ we have
$$L_g(s)=L_iota(s)L_varphi(s)=zeta(s+1)L_varphi(s).$$
Then by a Tauberian theorem, for example Wiener-Ikehara, we have
begineqnarray*
lim_ntoinftyfrac1n^2f(n)
&=&2operatornameRes(L_g,2)\
&=&2cdotoperatornameRes(zeta(s+1),2)cdotoperatornameRes(L_varphi,2)\
&=&2cdotzeta(3)cdotfrac12zeta(2)\
&=&fraczeta(3)zeta(2).
endeqnarray*

Your doubts are justified: $sum_d=1^n fraczeta(2)^-1d^2$ is just a sum of limiting probabilities, so it should (and does) yield $1$ in the limit. The sum you want to compute is of values of $frac1gcd(a,b)$, so you need to weight the possible values of that expression by their probabilities. This gives
$$
sum_d=1^n fraczeta(2)^-1d^2cdotfrac1dtofraczeta(3)zeta(2)approx 0.730763
$$
A numerical experimentat (with $n=1000$) gives a result close to $0.731$, so this seems likely to be correct.