Floor function and convergence of the sequence

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.