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.



An estimate on the summatory function of $\Big( \frac{n}{\phi(n)} \Big)^a$
1. Show that $$\left( \frac{n}{\phi(n)} \right)^a =\prod\_{p \mid n} \left(1 + \frac{1}{p-1} \right)^a \le \prod\_{p \mid n} \left(1 + \frac{C}{p} \right) = \sum\_{d \mid n} \frac{\mu(d)^2}{d} C^{\Omega(d)}$$
for some explicitly computable constant $C$ depending on $a$.
2. Bound your sum from above by
$$\sum\_{d \le x} \lfloor \frac{x}{d} \rfloor \left(\frac{\mu(d)^2}{d} C^{\Omega(d)} \right) \le x\sum\_{d \le x} \frac{C^{\Omega(d) }}{d^2}. $$
3. Finish by noting that $\Omega(d) = O(\frac{\ln d}{\ln \ln d})$ to conclude that $C^{\Omega(d)}$ grows slower than any power of $d$, hence $$\sum\_{d\ge 1} \frac{C^{\Omega(d) }}{d^2} < \infty.$$


