Suppose that $(n_k)_{k\in \mathbb{N}}$ is a given increasing sequence of positive integers.
Does there exist an (irrational) number $a$ such that $\{an_k\}:=(a n_k)\text{mod }1 \rightarrow 1/2$ as $k \rightarrow \infty$?
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3$\begingroup$ No, of course not in this generality: if $n_k=k$, then $ak$ will be dense in the unit circle (I assume you meant the fractional part of $an_k$, not $an_k$ itself). $\endgroup$– Christian RemlingCommented Feb 6, 2018 at 20:25
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$\begingroup$ Yes, sorry just edited it. $\endgroup$– John SmithCommented Feb 6, 2018 at 20:28
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$\begingroup$ On the other hand, if $n_k=10^{k^2}$, then $a=0.05005000050000005\dots$ should do. $\endgroup$– Gerry MyersonCommented Feb 6, 2018 at 22:50
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3$\begingroup$ There is an old theorem of Pollington (1979, Illinois J Math) showing that if $(n_k)$ is lacunary ($n_{k+1}/n_k\ge s$ for all $k$) for some $s>1$, then there exists a $\beta>0$ (depending on $s$) and an $a$ such that ${an_k}$ lies in $[\beta,1-\beta]$ for all $k$. A corollary of the proof is if $n_{k+1}/n_k\to\infty$, then there exists $a$ such that ${an_k}\to \frac 12$. $\endgroup$– Anthony QuasCommented Feb 7, 2018 at 3:54
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2 Answers
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The answer is no in general. For many increasing sequences $(n_k)_{k\in \mathbb{N}}$ of positive integers, it happens for every irrational number $a$ that $\{an_k\}$ is dense or even equidistributed in the unit circle. See this Wikipedia article for some examples.
answered Feb 6, 2018 at 22:17
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The sequence $n_{2k}=k^2, n_{2k+1}=k^2+1$ is a counterexample.
Indeed, if $\{ n_k a \} \to \frac{1}{2}$ then $\{ n_{2k}a \} \to \frac{1}{2}$ and $\{ n_{2k+1} a\} \to \frac{1}{2}$.
This implies that $a= n_{2k+1}a-n_{2k} a= \lfloor n_{2k+1}a\rfloor+ \{ n_{2k+1}a \} - \lfloor n_{2k}a\rfloor- \{ n_{2k}a \} $ and hence $$ a \pmod{1} \equiv \{ n_{2k+1}a \} - \{ n_{2k}a \} \to 0 \pmod{1} $$
This shows that $a \in \mathbb Z$, which contradicts $\{ n_k a \} \to \frac{1}{2}$.
The same is true about any subsequence containing infinitely many pairs of consecutive intgers.
P.S. On another hand, if $a$ is an irrational number, it follows from the denseness of $\{ na \}$ that there exists some $n_k$ such that $\{ n_ka \} \to \frac{1}{2}$.
This shows that there are many subsequences with this property, and I think one can argue that there are uncounatbly many such sequences.
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$\begingroup$ Why is $a(\text{mod} 1)\equiv \{n_{2k+1}a\}-\{n_{2k}a\}$? It could happen that $\{n_{2k+1}a\}-\{n_{2k}a\}<0$. $\endgroup$ Commented Feb 7, 2018 at 7:24
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$\begingroup$ @JohnSmith It is congruence $\pmod{1}$ not equality of fractional parts. $$a \equiv b \pmod{1}$$ means $a-b \in \mathbb Z$. Your point is exactly why I switched to $\pmod{1}$ there. $\endgroup$– Nick SCommented Feb 7, 2018 at 14:28