Edit: David Speyer's answer made me realize a couple of things and I would like to clarify. Sorry if the length of this is getting out of hand.
First, it is now clear that no estimate can be obtained for $\sigma(\alpha)$ without also obtaining an estimate for the irrationality measure of $\alpha$. This means that any technique used to estimate $\sigma(\alpha)$ can be translated into a technique for estimating the irrationality measure, and so it will not be any simpler as I'd hoped. Still, I wonder if there are any techniques that seem natural for estimating $\sigma(\alpha)$ (such as series transformations) that would not seem natural without introducing the series. For instance, though I'm sure it can be reasoned through directly, I would not have thought that the formula for the abscissa of convergence I referred to at the bottom in $(3)$ would provide an estimate for irrationality measure. Thus I suppose I'm still looking for a way to approach convergence of the series that does not involve simply obtaining a term-wise estimate (which would essentially be bounding the irrationality measure first). If you think this is a fool's errand, I'd certainly like to hear that as well.
Second (and this might merit its own question if I can't resolve it), in light of David's answer I wonder if anything more can be said about the relationship between $\sigma(\alpha)$ and the irrationality measure. Can they ever be different? If so: We know that the irrationality measure of an algebraic number is $2$; are there any examples where we can explicitly compute $\sigma(\alpha)$?
Finally, a minor correction to the definitions: What I've been calling $\mu(\alpha)$ is actually the irrationality measure of $\alpha$ minus one. I want to avoid getting caught up in trivial details so let me just redefine $\mu(\alpha)$ to be the least upper bound of the set of all exponents $s>0$ such that $\{n\alpha\} \geq n^{-s}$ for all but finitely many $s$. Everything below should remain the same.
Original Question
For $x\in\mathbb R$, let $\{x\} = x-\lfloor x\rfloor$ be the fractional part of $x$. I'm searching for an elementary proof that there exists a positive real number $s$ such that $$ \sum_{n = 1}^\infty {1\over n^s\{n\pi\}}<\infty. \tag{1} $$
The problem is certainly resolved by invoking the finite irrationality measure of $\pi$. It is known that $$ \{n\pi\} \geq {1\over n^8} \tag{2} $$ for all but finitely many integers $n$. (See the MathOverflow question numbers with known irrationality measure for references.) This implies that $(1)$ converges for $s>9$. But the proof of $(2)$ is very involved. There are other larger bounds on the irrationality measure of $\pi$ that have been known for longer and are probably easier to prove. But finite irrationality measure, while sufficient, does not seem necessary to ensure convergence of $(1)$. So I'm wondering if there isn't a qualitative way to prove that $(1)$ converges for some $s$ (no need to give an explicit $s$). While I say I seek an elementary proof, I'll settle for a proof that doesn't rely on explicit computation of irrationality measure.
The choice of $\pi$ is somewhat arbitrary—I made that choice to give a sense of concreteness to the problem. Perhaps a more general framework for the question is as follows. For irrational $\alpha$, define $\sigma(\alpha)$ to be the abscissa of convergence for the Dirichlet series
$$
\sum_{n = 1}^\infty {1\over n^s\{n\alpha\}}.
$$
Clearly $\sigma(\alpha)\geq 1$ in all circumstances. If $\alpha$ has irrationality measure $\mu(\alpha)$, then $\sigma(\alpha)\leq \mu(\alpha) + 1$ (since then $(2)$ holds with $\alpha$ in place of $\pi$ and $\mu(\alpha)+\epsilon$ in place of the exponent $8$ for any $\epsilon>0$). But it's conceivable that $\sigma(\alpha)$ could be much less than $\mu(\alpha)+1$. Indeed, this seems likely: If $s > 1$ and $n_1,n_2,n_3,\dots$ is a sequence of integers such that $\{n_i\alpha\}\leq n_i^{-s}$, then the numbers $n_i$ must grow very fast. [Edit: As David Speyer's answer indicates this is clearly foolish!]
Note: According to the Wikipedia page on general Dirichlet series, the abscissa $\sigma(\alpha)$ is given by \begin{align*} \sigma(\alpha) = \limsup_{n\to\infty}{\log{\left({1\over\{\alpha\}}+\cdots+{1\over\{n\alpha\}}\right)}\over\log{n}}.\tag{3} \end{align*} I just now found this formula, so I haven't yet played with it. Maybe it could help prove that $\sigma(\pi)$ is finite.