Is there an elementary proof that $\sum_{n=1}^\infty {1\over n^s\{n\pi\}}<\infty$ for some $s>0$?
It is not possible that $\sigma(\alpha)$ is much less than $\mu(\alpha)+1$. If $\sum \frac{1}{n^s \{n \alpha \}}$ converges, then $\lim_{n \to \infty} n^s \{n \alpha \} = \infty$. So, for all but finitely many $n$, we have $\{ n \alpha \} > n^{-s}$ and we deduce that $\mu(\alpha) \leq s$. Since $\sigma(\alpha)$ is the infimum of $s$ for which the sum converges, we get $\mu(\alpha) \leq \sigma(\alpha)$.
In summary, $$\mu(\alpha) \leq \sigma(\alpha) \leq \mu(\alpha)+1$$ and, in particular, one is finite if and only if the other is.