Let $\phi$ be Euler's totient function, and consider the sum: \begin{equation} \sum_{k=1}^{\infty} \left({1\over\phi(k)}\right)^s \end{equation} For what values of $s$ does this converge/diverge?
According to wolframalpha.com, it diverges even for $s = 100$:
(https://www.wolframalpha.com/input?i=sum%281%2F%28totient%28k%29%5E100%29%29)
I found that claim somewhat surprising, the intuition being that $\phi(k)$ is usually not horrendously smaller than $k$, so the sum above should converge if the real part of $s$ is not too small.
Using the product formula for $\phi$, we can give an Euler product expansion of the sum as: $$ \prod_p \left( 1 + \frac{1}{(p-1)^s} (1 + 1/p^s + 1/p^{2s} + \cdots )\right) $$ $$ = \prod_p \left( 1 + \frac{1}{(p-1)^s}\frac{1}{1 - 1/p^s} \right) $$ $$ = \prod_p \left( 1 + \left(\frac{p}{p-1}\right)^s \frac{1}{p^s - 1}\right) $$ with the product taken over primes. For real $s \ge 2$ this is bounded above by $$ \prod_p \left( 1 + 2^s \frac{2}{p^s} \right) $$ Comparing with $\sum_p \frac{1}{p^s}$, I suspect this is in fact convergent for $s \ge 2$, despite what wolframalpha says?
The sum converges if and only if $s>1$. (If we allow complex values, the sum converges if and only if the real part of $s$ is at least $1$.) The divergence of $\displaystyle\sum_{n=1}^\infty \frac1{\phi(n)^s}$ for $0<s\le1$ follows from the trivial comparison $\displaystyle\sum_{n=1}^\infty \frac1{\phi(n)^s} > \sum_{n=1}^\infty \frac1{n^s}$ (and the divergence for $s\le0$ is even more trivial).
For the complementary case, we want to use the fact that $\phi(n)$ is always pretty darn close to $n$: for example, it is known that $\displaystyle\lim_{n\to\infty} \frac{\phi(n)}{n^{1-\delta}} = \infty$ for any $\delta>0$. (Much stronger bounds are listed there, but this one suffices for us.) In particular, $\phi(n) > n^{1-\delta}$ for sufficiently large $n$, so by the comparison test, $\displaystyle\sum_{n=1}^\infty \frac1{\phi(n)^s} < \sum_{n=1}^\infty \frac1{n^{s(1-\delta)}}$ which converges whenever $s>\dfrac1{1-\delta}$. Since $\delta>0$ can be chosen arbitrarily, we conclude that $\displaystyle\sum_{n=1}^\infty \frac1{\phi(n)^s}$ converges whenever $s>1$.