Sequence $\{a(n)\}$ of real numbers is such that $\forall\space\lambda\in(1,2)$ sequence $a(\lfloor{\lambda}^n\rfloor)$ has a finite limit. Does it follow that $\{a(n)\}$ is convergent?
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1Define $S_\lambda = {\lfloor \lambda^m \rfloor \colon m\in\Bbb N}$. I believe the following lemma is true: there exists a fixed sequence ${b_m}$ of positive integers such that for all $\lambda\in(1,2)$, the intersection of sets ${b_m\colon m\in\Bbb N} \cap S_\lambda$ is finite. If true, the lemma would imply a negative answer to your question: take $a(n) = 1$ if $n\in {b_m\colon m\in\Bbb N}$ and $a(n)=0$ otherwise. – Greg Martin Dec 16 '15 at 07:47
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1My idea for constructing such a sequence ${b_m}$ (perhaps someone can reality-check it and flesh it out?): let $A_{c,k}$ be the set of all $\lambda\in(1+\frac1k,2)$ such that $c\in S_\lambda$. If $c$ is large then the measure of $A_{c,k}$ is small. Given $b_1,\dots,b_{j-1}$, if we choose candidates for $b_j$ at random from an interval of huge integers, then the expected measure of $$A_{b_j,j} \cap \bigg( \bigcup_{i<j} A_{b_i,j} \bigg)$$ should be small. Then some zero-one law should imply that with probability $1$, any given $\lambda$ is in only finitely many of the $A_{b_j,\infty}$...? – Greg Martin Dec 16 '15 at 07:54
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@GregMartin: Nice approach. I think I made it work the way you intended. The problem seems to me to be that we only get a sequence $(b_j) $ such that for almost all $\lambda $, we have what you want. I doubt that we can get a true "for all", since we are arguing using the Lebesgue measure on $(1,2) $, so that we "don't see" a null-set. – PhoemueX Dec 16 '15 at 10:50
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You're probably right. So are you saying it doesn't work at all, or that you found a variant that does? – Greg Martin Dec 16 '15 at 18:48
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@GregMartin: What I meant is that I made your approach work, so that I could show the "for almost all $\lambda $" statement. I do not know how/if one can obtain the "for all $\lambda $" statement (from this or otherwise). – PhoemueX Dec 17 '15 at 06:50
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The claim seems to be true if and only if the limit is independent of the choice of $\lambda$.
Lemma 1: Let $(m_n)$ be a strictly increasing sequence of natural numbers. There is a subsequence of $(m_n)$ that is also a subsequence of $\left\lfloor \lambda^n \right\rfloor$, for some $\lambda \in (1,2)$.
Proof: Note that $\left\lfloor \lambda^{s} \right\rfloor = m$ if and only if $$\lambda \in \left[ \exp\left(\frac{\log(m)}{s} \right),\exp\left(\frac{\log(m+1)}{s} \right) \right)=\left[ m^{1/s},(m+1)^{1/s} \right).$$ Note that the interval can be made arbitrarily thin by increasing $s$, since its length is $$(m+1)^{1/s}-m^{1/s} \leq 2^{1/s}-1.$$ By increasing $m$, the starting point of the interval can be chosen such that it's at least any $a \geq 1$. Combining these two steps, the interval of admissible $\lambda$'s can be fitted in any compact interval $[a,b]$ with $1 < a < b < 2$.
Let $$A_i = \left[ m_{t_i}^{1/s_i},(m_{t_i}+1)^{1/s_i} \right).$$ We show that the sequences $t_i$ and $s_i$ can be chosen so that $m_{t_i}$ equals the sequence $\left\lfloor \lambda^{s_i} \right\rfloor$ for some $\lambda \in (1,2)$. Choose $s_1$ and $m_{t_1}$ so that $A_1 \subset (1,2)$. Let $C_1$ be a compact interval in $A_1$. Choose $s_2$ and $m_2$ so that $A_2 \subset C_1$. Then choose a compact interval $C_2 \subset A_2$ and choose $s_3$ and $m_{t_3}$ similarly. Repeat these steps to obtain $A_k$ and $C_k$ for any $k$. Now $$\bigcap \limits_{k} A_k \supset \bigcap \limits_{k} C_k = C,$$ which is nonempty as an intersection of decreasing sequence of compact sets. Now $\lambda \in C$ is such that $$\lambda \in \left[ m_{t_i}^{1/s_i},(m_{t_i}+1)^{1/s_i} \right),$$ or equivalently $\left\lfloor \lambda^{s_i} \right\rfloor = m_{t_i}$ for all $i$. Hence, $m_{t_i}$ is a subsequence of $\left\lfloor \lambda^i \right\rfloor$.
Theorem 1: Assume that a sequence $a_n$ satisfies the hypothesis of the above question and that the limit does not depend on the choice of $\lambda$, then the sequence $a_n$ converges.
Proof: Let $a_{m_n}$ be any subsequence of $a_n$. Since $m_n$ is a strictly increasing sequence of natural numbers, by Lemma 1, there is a subsequence of $m_n$, denoted $m_{k_n}$, which is also a subsequence of $\left\lfloor \lambda^n \right\rfloor$, for some $\lambda \in (1,2)$. Since $a_{\left\lfloor \lambda^n \right\rfloor}$ converges, $a_{m_{k_n}}$ converges as its subsequence. Since a convergent subsequence can be extracted along an arbitrary subsequence of $a_n$, the entire sequence converges.
Matias Heikkilä
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