TITLE: What fraction of the integer lattice can be seen from the origin?
QUESTION [48 upvotes]: Consider the integer lattice points in the positive quadrant $Q$ of $\mathbb{Z}^2$.
Say that a point $(x,y)$ of $Q$ is visible from the origin if the
segment from $(0,0)$ to $(x,y) \in Q$ passes through no other point of $Q$.
So points block visibility, and the only points visible from the origin are those with
a clear line of sight:
   
Let $\nu(n)$ be the ratio of the number of points visible within the
square with corner $(n,n)$ to $n^2$. For example, for $n=8$, $43$ of the $64$ points
are visible, so $\nu(n) = 43/64 \approx 0.67$.

Q: What is $\lim_{n \to \infty} \nu(n)$?

It appears to be approaching $\approx 0.614$:
   
The question bears some similarity to Polya's Orchard Problem
(T.T. Allen, "Polya's orchard problem,"
The American Mathematical Monthly
93(2): 98-104 (1986). Jstor link), but I cannot see my question
is answered in that literature.
One could ask the same question for $\mathbb{Z}^d$.

Answered by Pete Clark: the limit is $\frac{1}{\zeta(d)}$. 
Thus in higher
dimensions, almost all the lattice points are visible (because $\zeta(d)$ approaches $1$).

REPLY [3 votes]: Here is an elementary argument using Euler products. I'm not sure if the initial probability part is rigorous, but it's a heuristic argument at least. The probability of a prime $p$ dividing a positive integer is $\frac{1}{p}$. Then the probability of $p$ dividing all entries of a $d$-tuple of positive integers is $\frac{1}{p^d}$. A $d$-tuple has $\gcd$ equal to $1$ if and only if no prime divides all of the elements. So the probability of a $d$-tuple of positive integers being coprime is $$\prod_{p\text{ prime}}{\left(1-\frac{1}{p^d}\right)}.$$ By the formula for an infinite geometric series,
$$\frac{1}{1-\frac{1}{p^d}}=\sum_{k=0}^{\infty}{\frac{1}{p^{kd}}},$$ so the above product is equal to $$\left[\prod_{p\text{ prime}}{\left(1+\frac{1}{p^{d}}+\frac{1}{p^{2d}}+\frac{1}{p^{3d}}+\cdots\right)}\right]^{-1}=\left[\sum_{n=1}^{\infty}{\frac{1}{n^d}}\right]^{-1}=\frac{1}{\zeta(d)}.$$