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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.

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  • $\begingroup$ Shouldn't equation (2) be strictly greater? $\{n\pi\}$ is irrational because $ir_{1}-r_{2}\in\mathbb{R}-\mathbb{Q}$ for all $i \in\mathbb{R}-\mathbb{Q}$ and $r_{1}, r_{2}\in\mathbb{Q}$. An irrational number cannot be equal to a rational one. $\endgroup$ – S V May 13 at 18:04
  • $\begingroup$ Who is Community? by definition having a username implying more that one user is responsible and not even specifying which community it represents is a dubious proposition at best $\endgroup$ – Adam Aug 27 at 3:31
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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.

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  • $\begingroup$ Ah that is a nice observation (that I feel foolish for missing)! I'm going to amend the question momentarily to reflect it. I wonder if this is the tightest relationship between $\mu$ and $\sigma$. And now I'm even more curious for a direct proof that the series converges (for, say, $\pi$), since that would be a proof of finite irrationality measure. $\endgroup$ – Nick Strehlke Jun 18 '14 at 1:19
  • $\begingroup$ Also, I think both of our $\mu$s are off (this is my fault from $(2)$, and I'll fix the question): The irrationality measure should be the infimum of $s$ such that $|\alpha - m/n| \leq n^{-s}$ has finitely many solutions $m,n$. To say that $\{n\alpha\}\leq n^{-s}$ has finitely many solutions is to say that $\alpha - \lfloor n\alpha\rfloor/n\leq n^{-s-1}$ has finitely many solutions, which (and this requires an argument but I'll take it on faith) is to say that $|\alpha - m/n|\leq n^{-s-1}$ has finitely many solutions. So $\mu\leq s+1$, not $s$. $\endgroup$ – Nick Strehlke Jun 18 '14 at 1:26
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Identifying $S^1$ with $[0, 1]$ with the boundary identified, the map $f: S^1 \to S^1$ defined by $f(x) = x + \pi$ is ergodic. That implies that the mean soujourn time of $f$ for the set $[0, \epsilon]$ is $1/\epsilon$. Maybe that could give an estimate on the sum strong enough to prove that it converges for suitable $s$? (I don't know whether ergodic theory satisifes the 'elementary' proof requirement, but there's no complex analysis involved.)

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  • $\begingroup$ I'll have to look into this tomorrow but it sounds like an interesting direction. I didn't really mean to impose an elementary requirement—what I really want is just a direct proof that the series converges that doesn't just boil down to computing irrationality measure in the usual way (if such a thing exists). $\endgroup$ – Nick Strehlke Jun 18 '14 at 1:32

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