Let $g: N→C$ be the arithmetical function giving the Dirichlet coeﬃcients of the Dirichlet series 1/ζ(2s). Prove that $|µ(n)| =\sum_{d|n} g(d) $ Let  $g: N→C$ be the arithmetical function giving the Dirichlet coeﬃcients of the Dirichlet series 1/ζ(2s).
a) Prove that g($k^2$) = µ(k) for every integer k, and that g(k) = 0 when k is not a square
b) Prove that $$|µ(n)| =\sum_{d|n} g(d) $$ Deduce that $$\sum_{n≤x} |µ(n)| = \sum_{nm≤x } g(m)=\sum_{nk^2≤x} |µ(k)|=\sum_{\sqrt k ≤x} |µ(k)|\sum_{n≤x/k^2} 1=\sum_{k ≤\sqrt x} µ(k) \lfloor\frac{x}{k^2}\rfloor$$ c) Use the previous equality to deduce that
#{n ∈N: n ≤ x, n is squarefree} =$\sum_{n≤x} |µ(n)|$= x/ζ(2) + O(√x).
d) Deduce that for every θ > 1/2 there exists $x_0 = x_0(θ)$ such that {n ∈N: n ∈ [x,x + $x^θ$], n is squarefree} is not empty ∀x ≥ $x_0$
I know this are a lot of question but for the last three I'm in trouble, the first I know that I can do it 
 A: a) First, one needs to find the Dirichlet series for $\frac{1}{\zeta(s)}$. Assuming $\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$, since $\zeta(s)\frac{1}{\zeta(s)} = 1$ from the formula for the coefficients of the product of two Dirichlet series one gets:
$\zeta(s)\frac{1}{\zeta(s)} = (\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}})(\sum_{n=1}^{\infty}\frac{1}{n^{s}})=\sum_{n=1}^{\infty}\frac{(a*1)_{n}}{n^{s}},$
where $1 = 1(n)$ is the identity arithmetic function and $(a*1)$ a Dirichlet convolution. From this relation, $a*1 = \epsilon$, where $\epsilon$ is the multiplicative identity. From Möbius inversion formula, $a = \epsilon*\mu$ where $\mu$ is the Möbius function. Hence $a_{n} = \mu(n), n = 1,2,3...$
From the findings above, $\frac{1}{\zeta(2s)} = \sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2s}} = \sum_{n=1}^{\infty}\frac{\mu(n)}{(n^{2})^{s}} = \sum_{n=1}^{\infty}\frac{g(n)}{n^{s}}$. From this last equality, one obtains that $g(n^{2})=\mu(n) \text{ for } n\in\mathbb{N}$ and $g(n) = 0 $ when $n$ is not the square of a natural number.
b) $g(n)$ is multiplicative:
Let $n,m \in \mathbb{N}$ be such that $(n,m) = 1$. Suppose that $nm$ is the square of some natural number. Given a prime $p$ such that $p|nm$, let $\alpha > 0$ be the largest exponent such that $p^{\alpha}|nm$. If $p|n$, then $p^{\alpha}|n$ otherwise $(n,m) \neq 1$. Since $2|\alpha$, from the fundamental theorem of arithmetic one gets that $n$ is the square of some natural number. It is also clear that $m$ is also a square. Therefore, $g(nm) = \mu(\sqrt{nm})=\mu(\sqrt{n})\mu(\sqrt{m}) = g(n)g(m)$. Now suppose $nm$ is not a square, then it is clear that $n \text{ and } m$ can not be both squares. Hence $g(nm) = g(n)g(m) = 0$.
Define $S(n) = \sum_{d|n}g(d)$. Because $g$ is multiplicative so is $S$, therefore one only needs to show that $S(p^{n})=|\mu(p^{n})|$, for powers of primes. It is clear that:
$$S(p^{n}) = \sum_{k = 0}^{n} g(p^{k}) = \sum_{k = 0}^{ \lfloor n/2 \rfloor}g(p^{2k}) = \sum_{k = 0}^{\lfloor n/2 \rfloor}\mu(p^{k}).$$
For $n = 1, \text{ since } \lfloor n/2 \rfloor = 0, \text{ then } S(p) = \mu(1) = |\mu(p)| = 1.$ For $n > 1$, since $\lfloor n/2 \rfloor > 0,$ then $S(p^{n}) = \mu(1) + \mu(p) = |\mu(p^{n})| = 0.$
Given arithmetic functions $f \text{ and } h$ such that $h(n) = \sum_{d|n}f(d)$, it is possible to show that:
$$\sum_{n \leq x}h(n) = \sum_{n \leq x}\lfloor \frac{x}{n} \rfloor f(n).$$
Applying the formula above, one gets:
$$\sum_{n \leq x}|\mu(n)| = \sum_{n \leq x}\lfloor \frac{x}{n} \rfloor g(n) = \sum_{k^{2} \leq x} \lfloor \frac{x}{k^{2}} \rfloor g(k^{2}) = \sum_{k^{2} \leq x} \lfloor \frac{x}{k^{2}} \rfloor \mu(k) = \sum_{k \leq \sqrt{x}} \lfloor \frac{x}{k^{2}} \rfloor \mu(k).$$
c) First one needs to show that:
$$\sum_{k > \sqrt{x}}\frac{x}{k^{2}} = O(\sqrt{x}).$$
Applying Euler's summation formula:
$$x\sum_{k > \sqrt{x}}\frac{1}{k^{2}} = x\int_{\sqrt{x}}^{\infty}\frac{1}{t^{2}}dt +  x\int_{\sqrt{x}}^{\infty} (t-\lfloor t\rfloor)(\frac{-1}{t^{3}})dt - \frac{x}{(\sqrt{x})^{2}}(\lfloor\sqrt{x}\rfloor - \sqrt{x}),$$
$$x\sum_{k > \sqrt{x}}\frac{1}{k^{2}} \leq x\int_{\sqrt{x}}^{\infty}\frac{1}{t^{2}}dt + x\int_{\sqrt{x}}^{\infty} |t-\lfloor t\rfloor|(\frac{1}{t^{3}})dt + \frac{x}{(\sqrt{x})^{2}}|\lfloor\sqrt{x}\rfloor - \sqrt{x}|,$$
$$x\sum_{k > \sqrt{x}}\frac{1}{k^{2}} \leq \sqrt{x} + \frac{1}{2} + 1.$$
Hence,
$$(\sum_{k \leq \sqrt{x}} \lfloor \frac{x}{k^{2}} \rfloor \mu(k))\zeta(2) = [\sum_{k \leq \sqrt{x}} (\lfloor \frac{x}{k^{2}} \rfloor - \frac{x}{k^{2}})\mu(k) + x\sum_{k = 1}^{\infty}\frac{\mu(k)}{k^{2}} - x\sum_{k > \sqrt{x}}\frac{\mu(k)}{k^{2}}]\zeta(2),$$
$$(\sum_{k \leq \sqrt{x}} \lfloor \frac{x}{k^{2}} \rfloor \mu(k))\zeta(2) = [\sum_{k \leq \sqrt{x}} (\lfloor \frac{x}{k^{2}} \rfloor - \frac{x}{k^{2}})\mu(k) - x\sum_{k > \sqrt{x}}\frac{\mu(k)}{k^{2}}]\zeta(2) + x,$$
$$|(\sum_{k \leq \sqrt{x}} \lfloor \frac{x}{k^{2}} \rfloor \mu(k))\zeta(2) - x| \leq [\sum_{k \leq \sqrt{x}} |\lfloor \frac{x}{k^{2}} \rfloor - \frac{x}{k^{2}}||\mu(k)| + x\sum_{k > \sqrt{x}}\frac{|\mu(k)|}{k^{2}}]\zeta(2),$$
$$|(\sum_{k \leq \sqrt{x}} \lfloor \frac{x}{k^{2}} \rfloor \mu(k))\zeta(2) - x| \leq (\sqrt{x} + \sqrt{x} + 1 + \frac{1}{2})\zeta(2).$$
Therefore,
$$|(\sum_{k \leq \sqrt{x}} \lfloor \frac{x}{k^{2}} \rfloor \mu(k)) - \frac{x}{\zeta(2)}| \leq 2\sqrt{x} + \frac{3}{2}.$$
d) $\#\{n \in \mathbb{N} : n \in [x, x + x^{\theta}], n \text{ is squarefree}\} = \sum_{ x \leq n \leq x + x^{\theta} } |\mu(n)| \geq \sum_{n \leq x + x^{\theta}}|\mu(n)| - \sum_{n \leq x}|\mu(n)| = \frac{x^{\theta}}{\zeta(2)} + O(\sqrt{x+x^{\theta}}) + O(\sqrt{x}).$
For any $\theta > 1/2$, it is easy to see that: 
$$lim_{x \rightarrow \infty}[\frac{x^{\theta}}{\zeta(2)} + O(\sqrt{x+x^{\theta}}) + O(\sqrt{x})] = +\infty.$$
Therefore, there exists $x_{0} = x_{0}(\theta)$ such that for $x \geq x_{0}$ then:
$$\frac{x^{\theta}}{\zeta(2)} + O(\sqrt{x+x^{\theta}}) + O(\sqrt{x}) \geq 1.$$
