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Let $x_n$ be a non-negative valued sequence and suppose that the following hold:

  • $\lim\limits_{n\to\infty} x_n =0$
  • There exists some polynomial function $p$ of degree at-least $1$ such that: $$ \|x_n\| \leq p(n^{-1}); $$ (not that there is no assumption that $p(0)=0$!)
  • $x_n\neq 0$ for any $n>0$.

Can we deduce the existence of a rational function $q$ and some positive integer $N$ satosfying:

  1. $q(0)=0$
  2. $\|x_n\|\leq q(n^{-1})$ for all $n\geq N$.
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    $\begingroup$ llama is spelled with two l's. $\endgroup$ Commented Nov 13, 2020 at 11:57
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    $\begingroup$ Lama is my name ; see: en.wikipedia.org/wiki/Lama_(name) $\endgroup$
    – ABIM
    Commented Nov 13, 2020 at 12:01
  • $\begingroup$ ok. your question doesn't make sense though. $p$ always exists and so does $q$. E.g. $q(n) = Cn$ for $C > \max_n x_n$ (which exists). $\endgroup$ Commented Nov 13, 2020 at 12:02
  • $\begingroup$ Good eye, it should be a polynomial in the reciprocal $\frac1{n}$. Now it should be fine. $\endgroup$
    – ABIM
    Commented Nov 13, 2020 at 12:04
  • $\begingroup$ I don't think so. $p$ still always exists. Take $p(t) = t+C$ for a large $C$. $\endgroup$ Commented Nov 13, 2020 at 12:06

1 Answer 1

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The answer is no (assuming that $\|x\|:=x$ for real $x\ge0$). Indeed, as noted by mathworker21, your condition on the existence of $p$ is always satisfied, by choosing, e.g., $p(t)=c+t$ with $c:=\sup_n x_n$.

On the other hand, your condition on the existence of $q$ will not hold e.g. if $x_n=1/\sqrt n$ for all natural $n$. Indeed, suppose the contrary: that there is some rational function $q$ such that $q(0)=0$ and $1/\sqrt n\le q(1/n)$ for all large enough $n$. Then $$q(t)=t^k\frac{P(t)}{R(t)}$$ for some integer $k\ge1$, some polynomials $P$ and $R$ with $R(0)\ne0$, and all $t$ in a neighborhood of $0$. So, $q(1/n)=O(1/n^k)=o(1/\sqrt n)$, which contradicts the condition that $1/\sqrt n\le q(1/n)$ for all large enough $n$.

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