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I am reading some materials on hypergeometric functions and their relation to general L-functions and the authors make use of the following "well-known elementary result from diophantine approximation":

Let $\lambda$ be a positive real number such that $\lambda\neq\frac{1}{2m}$ for all $m \in\mathbb{N}$. Then $\left\{ \frac{k}{2\lambda} \right\}$ does not converge when $k\to\infty$, where $\{ \cdot \}$ denotes the fractional part of a real number.

As I have no idea of Diophantine Approximation yet, I would like to ask why this fact is true. Is there an easy way to see its validity? I would be also thankful for references related to this statement.

Thanks in advance, efq

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up vote 7 down vote accepted

Let $x$ be a fixed positive real. Then, the behavior of the sequence $\{ kx \}$ falls under one of the following three types, depending on $x$.

  1. If $x$ is an integer: In this case, the sequence $\{ kx \}$ is the constant sequence $0$. In particular, the sequence converges to $0$.

  2. If $x$ is rational, but not an integer: Suppose $x = a/b$ where $a,b$ are positive integers with $\gcd(a,b)=1$, then the sequence is periodic with period $b$. Moreover, $kx = 0$ iff $k$ is divisible by $b$, and hence it is not a constant sequence. It then follows that $\{ kx \}$ does not converge as $k \to \infty$. (In fact, the sequence has $b$ limit points, namely $\lbrace 0, \frac{1}{b}, \frac{2}{b}, \ldots, \frac{b-1}{b} \rbrace$.)

  3. If $x$ is irrational: In this case, the equidistribution theorem states that the sequence $\{ kx \}$ is uniformly distributed over the unit interval. In particular, the set $$ A := \{ \{kx \} \,:\, k \in \mathbb N \} \tag{1}$$ is dense in $[0,1]$, and thus the sequence cannot converge. I will also give a more elementary proof of this claim below.

Now, in your case, applying $x= \frac{1}{2\lambda}$ gives the claim made by the authors.

An argument using Dirichlet's approximation theorem. The equidistribution theorem seems like an overkill for this question. Here I give an alternate argument via Dirichlet's approximation theorem to show that the set $A$ (as in $(1)$) is dense in the interval $[0,1]$, and hence the sequence in question cannot converge. I think this is what the authors had in mind.

Dirichlet's theorem says that for any irrational $x$, there are infinitely many $q$ such that $\{ qx \} < \frac{1}{q}$. In particular, for any $\varepsilon>0$, there exists a positive integer $q$ such that $\{ qx \} < \varepsilon$. (Already this shows that $0$ is a limit point of the set $A$.)

Now fix an interval $(a,b) \subseteq [0,1]$ and let $0 < \varepsilon < b-a$. Then we know that there exists $q$ such that $\{ qx \} < \varepsilon$. Denoting $\{ q x \}$ by $\delta$, we have $0 < \delta < \varepsilon$. Now, consider the sequence $$ \{ qx \}, \{2qx \}, \{3qx\}, \ldots, \{n qx \}, \ldots $$ It is easy to see that $\{ nqx \} = n\delta$ for $n < \frac{1}{\delta} $. In other words, the sequence jumps by a distance of $\delta > 0 $ every time, until it exceeds $1$ (and hence it wraps around). However, since the length of the interval $(a,b)$ is larger than $\varepsilon > \delta$, the sequence must land in the interval $(a,b)$ for some $n$. (More precisely, for $$n = \left\lfloor \frac{a}{\delta} \right\rfloor + 1 = \left\lfloor \frac{a+\delta}{\delta} \right \rfloor ,$$ we have $\{ qnx \} \in (a,b)$. I will leave this last claim as an exercise.)

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Thank you for your answer! With the above references and your explanations, I believe I will be able to work my way through to the result in question. In particular, the addendum looks like the thing I have been after since it does not use any heavy machinery (which presumably was the intention of the authors, too). I appreciate you taking the time to add it! – ex-falso-quodlibet Sep 26 '11 at 4:34

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