Asymptotic formula for $n^{th}$ square-free I face an exercise : Find asymptotic formular for the $n^{th}$ square-free number with the error $O(n^{1/2}).$
I don't quite understand the question. What is the $n^{th}$ square-free number ? I guess it is primorial ? Like $2$ is the first, $2 \times 3$ second, $2 \times 3 \times 5$ third , ... (although I do not sure if it includes the number like $2 \times 5$ since $3$ is missing, and if it includes, what should be the third square-free number ? $2 \times 3 \times 5$ or $2 \times 5$.)
Let assume that I want to find an asymptotic formula for $p_1...p_k$ where $p_k$ is the $k^{th}$ prime. So I need to find $g$ such that $$\lim_{x \rightarrow \infty} \frac{\prod_{p \leq x} p}{g(x)} = 1.$$
Basically, asymptotic $\sim$ can be related to little $o$, but this question mention something about error term $O(n^{1/2})$. How can asymptotic relate to big $O$ ?
That makes me confused.
For the method, I see that the primorial can be related to $e^{\theta(x)}$, where $\theta$ is the Chebyshev function. But I do not know how to do $\sim$ with big $O(n^{1/2}).$
 A: By definition of the Möbius function we have that $n$ is squarefree iff $\mu^2(n)=1$. Thefore, the function $\eta(x)=\sum_{n\leq x}\mu(n)^2$ counts the number of squarefree integers less or equal to $x$. Now recall the following elementary property of $\mu^2$:
$$\mu^2(n)=\sum_{d^2|n}\mu(d),$$
where the sum is extended over all the divisors $d$ of $n$ such that $d^2|n$. Thus,
$$\eta(x)=\sum_{n\leq x}\mu^2(n)=\sum_{n\leq x}\sum_{d^2|n}\mu(d).$$
Write $n=d^2q$. Then $n\leq x$ iff $d^2\leq x$ (i.e., $d\leq\sqrt x$) and $q\leq x/d^2$, so
$$\sum_{n\leq x}\sum_{d^2|n}\mu(d)=\sum_{d\leq\sqrt x}\sum_{q\leq x/d^2}\mu(d)=\sum_{d\leq\sqrt x}\mu(d)\left(\frac x{d^2}+O(1)\right)
=x\sum_{d\geq1}\frac{\mu(d)}{d^2}-x\sum_{d\geq\sqrt x}\frac{\mu(d)}{d^2}+O(\sqrt x).$$
But
$$\sum_{d\geq\sqrt{x}}\frac{\mu(d)}{d^2}=O\left(\sum_{d\geq\sqrt x}\frac1{d^2}\right)=O\left(\frac1{\sqrt{x}}\right)$$
so
$$x\sum_{d\geq1}\frac{\mu(d)}{d^2}-x\sum_{d\geq\sqrt x}\frac{\mu(d)}{d^2}+O(\sqrt x)
=x\sum_{d\geq1}\frac{\mu(d)}{d^2}+O(\sqrt{x}).$$
Finally, since $\sum_{d\geq1}\frac{\mu(d)}{d^2}=1/\zeta(2)$, we conclude that
$$\eta(x)=\frac{x}{\zeta(2)}+O(\sqrt x)=\frac{6x}{\pi^2}+O(\sqrt x).\tag{$*$}\label{eq:1}$$
EDIT: from the above we have that $\eta(x)\sim \frac{x}{\zeta(2)}$, because
$$\eta(x)=\frac{x}{\zeta(2)}+O(\sqrt x)=\frac{6x}{\pi^2}+o(x)=\frac{6x}{\pi^2}(1+o(1))$$
since $\sqrt{x}=o(x)$ (i.e., $\lim_{x\to\infty}\sqrt{x}/x=0$.) Nevertheless, we have proved something better. Let us write \eqref{eq:1} in the following equivalent form:
$$\frac{\eta(x)}{x/\zeta(2)}=1+O\left(\frac1{\sqrt{x}}\right).$$
We see that the function $\frac{\eta(x)}{x/\zeta(2)}$ approaches to $1$ at least as fast as $\frac1{\sqrt{x}}$ approaches to zero. This provides us more information than simply saying that $\frac{\eta(x)}{x/\zeta(2)}$ converges to $1$. For example, consider the question "What is the value of  $\lim_{x\to\infty}\sqrt[3]{x}(\frac{\eta(x)}{x/\zeta(2)}-1)$?" You can answer it if you know that $\frac{\eta(x)}{x/\zeta(2)}-1=O(1/\sqrt{x})$.
A: Hint:
$\mu(n)^2$, where $\mu(n)$ is the Moebius function, is the characteristic function of square-free numbers.
By convolution of Dirichlet series and summation by parts, it is not difficult to show that
$$ \sum_{n\leq x}\mu(n)^2 \approx \frac{x}{\zeta(2)} = \frac{6x}{\pi^2} \tag{1}$$
i.e. the square-free integers form a subset of the natural numbers with positive density. By $(1)$, it follows that the $n$-th square-free natural number is close to $\color{red}{\frac{\pi^2 n}{6}}$. How close depends on a explicit error term for $(1)$.
A: A square-free number is one with no square divisors (other than 1).
They're not the same as primorial numbers: 3 is square-free but not primorial. Here's a list of the first square-free numbers, at the On-Line Encyclopedia of Integer Sequences.
Does that get you unstuck?
