Tail estimates for the sum $\sum_{n\geq 1} (1/ \varphi(n)^2)$ I've been trying to prove the following fact: If $\varphi$ denotes Euler's totient function, then for $Q\geq 1$ we have
\begin{align*}
\sum_{n\geq Q} \frac{1}{\varphi(n)^2} \ll Q^{-1}.
\end{align*}
I am able to prove that $\sum_{n\geq Q} \frac{1}{n^2} \ll Q^{-1},$ and I'm familiar with Euler products but unclear on the best way to apply them here. I am also aware of the estimate $\varphi(n)\gg n/\log (\log(n))$. Any hints would be really helpful! Do I need to use some kind of averaging argument for $\varphi?$
 A: One method that often works in situations like this is approximating the function in question by an easier to handle function. For most $n$, $\varphi(n)$ is "close" to $n$, so it makes sense to approximate $1/\varphi(n)^2$ by $1/n^2$. Of course we must then somehow get a grip on the error introduced by that approximation. This can often be achieved by writing the function one is interested in as the Dirichlet convolution of the simpler function and a function that is "small". Thus let us define $f$ by
$$\frac{1}{\varphi^2} = \frac{1}{\operatorname{id}^2} \ast f\,, \quad \text{i.e.} \quad f = \frac{\mu}{\operatorname{id}^2} \ast \frac{1}{\varphi^2}\,.$$
Since everything is multiplicative we only need to consider prime powers. We find
$$f(p) = \frac{1}{1}\cdot \frac{1}{(p-1)^2} + \frac{-1}{p^2}\cdot \frac{1}{1^2} = \frac{2p-1}{p^2(p-1)^2}$$
and
$$f(p^k) = \frac{1}{1}\cdot \frac{1}{(p-1)^2p^{2k-2}} + \frac{-1}{p^2}\cdot \frac{1}{(p-1)^2p^{2k-4}} = 0$$
for $k \geqslant 2$. Thus $f$ is supported on the squarefree numbers, and
$$f(n) = \lvert\mu(n)\rvert \prod_{p \mid n} \frac{2p-1}{p^2(p-1)^2} = \frac{\lvert \mu(n)\rvert}{n^2\varphi(n)^2}\prod_{p \mid n} (2p-1)\,.$$
We have the bounds $0 \leqslant f(n) \leqslant \frac{2^{\omega(n)}}{n\varphi(n)^2}$, which show that $\sum n^{\alpha}f(n)$ converges for all $\alpha < 2$. So $f$ is indeed nicely small. Now we rewrite our tail sum:
\begin{align}
\sum_{n \geqslant Q} \frac{1}{\varphi(n)^2}
&= \sum_{km \geqslant Q} \frac{1}{m^2}f(k) \\
&= \sum_{k \leqslant Q} f(k)\sum_{m \geqslant Q/k} \frac{1}{m^2} + \sum_{k > Q} f(k) \sum_{m = 1}^{\infty} \frac{1}{m^2}\,.
\end{align}
Using $\sum_{m \geqslant y} 1/m^2 \leqslant 2/y$ for all $y \geqslant 1$ we obtain
$$\sum_{k \leqslant Q} f(k)\sum_{m \geqslant Q/k} \frac{1}{m^2} \leqslant \frac{2}{Q} \sum_{k \leqslant Q} kf(k) < \frac{2}{Q} \sum_{k = 1}^{\infty} kf(k)\,.$$
The missing estimate $\sum_{k > Q} f(k) \ll Q^{-1}$ follows easily from the above bound, which via the lower bound for $\varphi(n)$ and $\omega(n) \ll \frac{\log n}{\log \log n}$ implies $f(n) \ll n^{\varepsilon - 3}$ (and thus $\sum_{n > Q} f(n) \ll Q^{\varepsilon-2}$) for all $\varepsilon > 0$. It can also be obtained via summation by parts from the fact that $\sum nf(n)$ converges.
Altogether, this shows the desired
$$\sum_{n \geqslant Q} \frac{1}{\varphi(n)^2} \ll Q^{-1}\,.$$
Although $1/\varphi(n)^2$ can be much larger than $1/n^2$, this happens rarely enough that the tail sums have the same order of decay.
