A better estimate on the summatory function of $\Big( \frac{n}{\phi(n)} \Big)^a$

In this question, the OP asks:

Prove that for given $a>0$ there exists $C(a)$ depending on $a$ such that $$\sum_{n\leq x}\Big(\frac{n}{\phi{(n)}}\Big)^a\leq C(a)x$$where $\phi(n)$ denotes Euler totient function.

The answers and comments point out how to show the upper bound, but experiment shows that the growth (for any given $a$) is, in fact, linear - here is a graph for $a=1:$ How would one show this?

• Morally speaking each prime $p$ contributes a factor of about $1 + (a/p^2)$ to the average of $(n/\phi(n))^a$, and the product of those factors converges by comparison with $\zeta(2)^a$. You can probably prove this by setting up the Dirichlet series for $(n/\phi(n))^a$ and showing it has a simple pole at $s=1$ and no other poles in some half-plane containing $1$. – Noam D. Elkies Sep 30 '16 at 21:19

1. Write $\left( \frac{n}{\phi(n)} \right)^a$ as $\prod_{p \mid n}(1+g(p))=\sum_{d \mid n} g(d)\mu^2(d)$, where $g$ is completely multiplicative and given by $g(p)=\left(1 + \frac{1}{p-1} \right)^a - 1$. Note that $$g(p) \le \frac{2^a-1}{p-1},\quad g(d) \le \frac{\mu^2(d)}{d} (2^{a+1}-2)^{\Omega(d)}.$$
2. The sum becomes $$\sum_{d \le x} \lfloor \frac{x}{d} \rfloor g(d)\mu^2(d) =x\sum_{d \le x} \frac{g(d)\mu^2(d)}{d} + O\left( \sum_{d\le x} g(d)\mu^2(d) \right)$$ $$\qquad =x\sum_{d} \frac{g(d)\mu^2(d)}{d} + O\left( \sum_{d\le x} g(d)\mu^2(d) \right) + O\left( x\sum_{d>x} \frac{g(d)\mu^2(d)}{d} \right).$$
3. Finish by noting that $\sum_{d\le x} g(d)\mu^2(d)$ is $o(x)$, and that $\sum_{d>x } \frac{g(d)\mu^2(d)}{d}$ is $o(1)$ (so in particular $C(a):=\sum_{d } \frac{g(d)\mu^2(d)}{d}=\prod_{p} \left( 1+\frac{g(p)}{p}\right)$ converges): $$\Omega(d) = O(\frac{\ln d}{\ln \ln d}) \implies$$ $$\sum_{d\le x} g(d)\mu^2(d) \le \sum_{d\le x} \frac{(2^{a+1}-2)^{O(\frac{\ln d}{\ln \ln d})}}{d} =\sum_{d\le x} O_a\left( \frac{1}{\sqrt{d}} \right) = O_a(\sqrt{x}),$$ $$\sum_{d>x} \frac{g(d)\mu^2(d)}{d} \le \sum_{d>x} \frac{(2^{a+1}-2)^{O(\frac{\ln d}{\ln \ln d})}}{d^2} =\sum_{d>x} O_a\left( \frac{1}{d\sqrt{d}} \right) = O_a(\frac{1}{\sqrt{x}}).$$
Remark: The estimate $\Omega(d) = O\left( \frac{\ln d}{\ln \ln d} \right)$ is elementary - it relies only on the Chebyshev estimates. In fact, one only needs $\Omega(d) = o(\ln d)$, which follows from the fact that $\pi(d) = o(d)$ (zero-density of primes), a classical result already due to Legendre.