I have to show that if $\phi$ is Euler's totient function, then the series $\sum\limits_{n=2}^{\infty} \frac{1}{\phi (n) \log n}$ diverges and $\sum\limits_{n=2}^{\infty} \frac{1}{\phi(n) \log^2 n}$ converges, but there's something I don't understand.
We know both $\phi$ and $\log$ go off to infinity. So shouldn't they both converge?
And what about the series: $\sum\limits_{n=2}^{\infty} \frac{1}{\pi (n) \log n}$ and $\sum\limits_{n=2}^{\infty} \frac{1}{\pi (n) \log^2 n}$, where $\pi$ is the function that counts the prime numbers up to n?
It is easy to check that the first does not converge using that $\pi (n) \leq C \frac{n}{log n}$ for some positive constant $C$ but $\pi (n)$ also goes off to infinity. So is there a result analogue to this that would give the desired result for $\phi$?