Vanishing sequence and subsequence with particular decay [closed]

Question

Assume I have a sequence $\{a_m\}$ that is vanishing and strictly positive: $$ 0<a_{m+1}\leq a_m\leq\ldots\leq a_1<\infty, \quad \lim_{m\to \infty}a_m = 0 $$ Is it true or false that this has a subsequence $\{a_{m_n}\}$ such that $$ a_{m_n}\asymp n^{-1} $$ which is to say, there exists a constant $C>0$ such that for all $n$: $$ \frac{1}{Cn} \leq a_{m_n} \leq \frac{C}{n} $$

Answer 1

The answer is no. E.g., let $a_m:=1/(m!)^2$ and $n_k:=k!(k+1)!$ for natural $m,k$. Your desired condition means that $|\ln na_{m_n}|$ is bounded. In our case, we have $$a_{k+1}<\frac1{n_k}<a_k$$ for all $k$. So, for all $m\le k$ we have $1<n_k a_k\le n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_k)|=\ln(k+1)$. Similarly, for all $m\ge k+1$ we have $1>n_k a_{k+1}\ge n_k a_m$ and hence $|\ln(n_k a_m)|\ge|\ln(n_k a_{k+1})|=\ln(k+1)$. So, $|\ln(n_k a_m)|\ge\ln(k+1)$ for all $k,m$.

So, for any $(m_n)$ $$|\ln n_ka_{m_{n_k}}|\ge\inf_m|\ln n_k a_m|\ge\ln(k+1)\to\infty$$ as $k\to\infty$, so that $|\ln na_{m_n}|$ is unbounded, for any choice of $(m_n)$.