Tail estimates for the sum $\sum_{n\geq 1} (1/ \varphi(n)^2)$

Question

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 $\varphi?$

Answer 1

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.