Asymptotic estimate of the sum $\sum_{n\leq x}1/\phi^2(n)$ How to show that we have the following estimate:
$$\sum_{n\leq x}\frac{1}{\phi^2(n)}=c+O(\frac{1}{x}),$$
where $\phi$ is the Euler's totient function and $c$ is a constant. I tried to use the convolution method, but I don't know how to choose the appropriate functions.
 A: $\phi(n)$ is multiplicative and is never $0$, hence $\frac{1}{\phi^2(n)}$ is multiplicative :
$$\frac{1}{\phi(p^k)^2} = \frac{1}{(p-1)^2 p^{2(k-1)}} = \frac{p^2}{(p-1)^2} p^{-2k}$$
hence :
$$F(s) = \sum_{n=1}^\infty \frac{1}{\phi(n)^2}n^{-s} = \prod_p \left(1 + \sum_{k=1}^\infty \frac{1}{\phi(p^k)^2}p^{-sk}\right) = \prod_p \left(1 + \frac{p^2}{(p-1)^2}\sum_{k=1}^\infty p^{-(s+2)k}\right)$$
and since when $p \to \infty$ : $\frac{p^2}{(p-1)^2}  = 1 + \mathcal{O}(p^{-1})$  we find that 
$$F(s) = \prod_p \left(1 + p^{-(s+2)} + \mathcal{O}(p^{-(s+3)})\right)$$
which converges absolutely for $Re(s +2) > 1$ i.e. for $Re(s) > -1$, hence, by a general theorem on the speed of convergence of Dirichlet series :
$$\sum_{n \le x} \frac{1}{\phi(n)^2} = \mathcal{O}(x^{-1+\epsilon}) + F(0) \qquad\qquad \forall \ \epsilon > 0$$
to prove that $\sum_{n \le x} \frac{1}{\phi(n)^2} = \mathcal{O}(x^{-1+\epsilon}) + F(0)$ we need a further step :
$F(s)$ has a simple pole at $s = -1$ and no other pole on $Re(s) = -1$ (because it is similar to $\zeta(s)$ on $Re(s) = 1$) hence it is true that :
$$\sum_{n \le x} \frac{1}{\phi(n)^2} = \mathcal{O}(x^{-1}) + F(0)$$
