What's equal this :$\sum_{\phi(n)=1}^{\infty}\frac{1}{\phi(n)}$?

Question

I'm interesting to know something about convergence of series related to the Euler totient functin, then my question here is :

Question

Is this a convergent sum :$\sum_{\phi(n)=1}^{\infty}\frac{1}{\phi(n)}$ ?

Note: $\phi$ is Euler totient function

Answer 1

I reckon $\phi(n)\le n$ and so $$\sum_{n=1}^\infty \frac1{\phi(n)}\ge\sum_{n=1}^\infty\frac1n$$ etc.

ADDED IN EDIT

I also reckon that if $A=\{\phi(n):n\in\Bbb N\}$ then $$\sum_{m\in A}\frac1m\ge\sum_p\frac1{\phi(p)}>\sum_p\frac1p$$ where $p$ runs through all primes.

Answer 2

Let $A=\{\,\phi(n)\mid n\in\Bbb N, n>1\,\}$. If $p$ is prime then in particular $\phi(p)=p-1\in A$. Hence $$\sum_{a\in A}\frac 1a\ge \sum_{p\text{ prime}}\frac1{p-1}\ge \sum_{p\text{ prime}}\frac1{p}=\infty. $$