Using the generating function $\zeta(s)/\zeta(2s)$ for the squarefree integers,how do you get the density result $6/\pi^2$? While I can see that $\zeta(s)/\zeta(2s)$ generates the squarefree integers, I cannot see how to get from this to the density result $6/\pi^2$. I know residue at s=1 is $6/\pi^2$ (why?) .A post density of squarefree numbers in $\mathbb{Z}$ that are 1 mod 4
 from January 4 2016 implies that this should be easy. On the way to answering a harder question, Ram Murty said: 
The first thing to note is that the density of odd squarefree numbers is $4/\pi^2$
 which is easily seen from the Dirichlet generating series for odd squarefree numbers
with the 2-Euler factor removed, from which the residue at s=1 is determined as $4/\pi^2$
My confusions are:
1 Not sure if odd squarefree is a distraction or relevant. Can we use the $6/\pi^2$ result directly?
2 The generating function seems to generate the sum of reciprocals to power s and not something related to a count. 
3  Why is  the residue at s=1 relevant, why not the value as s tends to zero, to get a count rather than the sum of reciprocals to power s? But then s and 2s both tend to zero and we get  $\zeta(0)/\zeta(0)$ which is 1 which is wrong.
4. To get a density don't we need a count of squarefree numbers less than n, divided by n. Then we want to show that this tends to $6/\pi^2$ as n tends to infinity?
5. But the Euler product expression $\zeta(s)/\zeta(2s)$ up to nth prime does not give all the squarefree up to the product of the first n primes. It only gives squarefree numbers attainable by multiplying combinations of first k primes. For example 2,3,5,7 multiplied gives 210 and there are lots of square free between 7 and 210. And I can't see how to make progress on an expression for density from which a limit for large n gives the correct result.
Can you help me resolve my confusions?
 A: If we start with $\mathbb{N}^+$ and perform a sieve by removing all the multiples of $2^2,3^2,5^2,\ldots,p^2$, by the inclusion-exclusion principle we get a set with density
$$ \left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{5^2}\right)\cdot\ldots\cdot\left(1-\frac{1}{p^2}\right)$$
so, with some care in managing the error terms (the intersection of nested sets, all of them with natural density, may not have a natural density) we have that the density of square-free numbers is given by
$$ \prod_{p}\left(1-\frac{1}{p^2}\right) $$
where Euler's product for the $\zeta$-function gives that
$$ \prod_{p}\left(1-\frac{1}{p^s}\right)^{-1} = \zeta(s) $$
holds for any $s>1$. Taking $s=2$ we have that the density of square-free numbers is $\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$.
For the same reason the density of the square-free numbers among the odd integers is given by
$$ \left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{5^2}\right)\cdot\ldots = \frac{4}{3}\cdot\frac{1}{\zeta(2)} = \frac{8}{\pi^2}$$
so the density of the odd square-free numbers is $\frac{4}{\pi^2}$.
