Define $$F(s) = \sum_{n=1}^\infty \frac{\mu^2(n)}{n^s}$$ where $s \in \mathbb{C}.$
Verify that "If $k$ is a constant and $g : \mathbb{N} \rightarrow \mathbb{C}$ with $g(n) = O_\delta(n^{\delta + k})$ for any $\delta > 0$, then the abscissa of absolute convergence, $\sigma_2(G)$, for $$G(s) = \sum_{n=1}^\infty \frac{g(n)}{n^s}$$ satisfies $\sigma_2(G) \leq k+1.$"
Then use the statement to show that $\sigma_1(F) = \sigma_2(F) =1.$
Note : Let $H(s)$ denote the Dirichlet series, then $$\sigma_1(H) = \inf\{\sigma \in \mathbb{R} : H(\sigma) \ \mbox{converges}\},$$ $$\sigma_2(H) = \inf\{\sigma \in \mathbb{R} : H(\sigma) \ \mbox{converges absolutely}\}.$$
$\textbf{Proof}$ Fix $\delta > 0$, then there exists $n(\delta)$ and a constant $c(\delta)$ such that $$|g(n)| \leq c(\delta)n^{k + \delta}$$ for any $n \geq n(\delta).$ So $$\sum_{n=n(\delta)}^\infty \frac{|g(n)|}{n^\sigma} \leq c(\delta) \sum_{n \geq n(\delta)}\frac{1}{n^{\sigma - k - \delta}}.$$ The last summation will be convergent if and only if $$\sigma > k + \delta + 1.$$ By definition, $$\sigma_2(G) \leq k + \delta + 1$$ for any $\delta > 0$.
Therefore $\sigma_2(G) \leq k+1$
$\textbf{Is my proof correct ?}$
$\textbf{Proof}$ Since $\frac{\mu^2(n)}{n^\sigma} \geq 0$ for any $\sigma \in \mathbb{R}$, $\sigma_1(F) = \sigma_2(F).$ Since $n^x$ is an incresing function and $n^x \rightarrow 1$ as $x \rightarrow 0$, $$\mu^2(n) = O_\delta(n^\delta)$$ for any $\delta > 0.$ Then $\sigma_2(F) \leq 1$. Since $$\sum_{p \ \mbox{prime}}\frac{1}{p} \leq F(1)$$ and $\sum_{p \ \mbox{prime}}\frac{1}{p}$ diverges, $\sigma_2(F) \geq 1.$