Let $(x\_n)\_{n\in\mathbb{N}}$ be a non-increasing sequence in [0,1], (i.e. $x\_n\ge x\_{n+1},n\in\mathbb{N} $), such that $x\_n\ge\frac{1}{n},n\in\mathbb{N} $.


If we fix $k\in\mathbb{N}$ is there necessarily a lower bound c>0 for the fractions $\frac{x\_{kn}}{x\_n}$ for all $n\in\mathbb{N}$?


I was thinking that if it is true then maybe $\frac{x\_{kn}}{x\_n}\ge\frac{1}{k}$.



Lower bound for $\frac{x_{kn}}{x_n}$, where $(x_n)_{n\in\mathbb{N}}$ is a non-increasing sequence in [0,1] with $x_n\ge\frac{1}{n}$
Such a lower bound does not exist in general.


E.g., for natural $j$ and natural $n\in((j-1)!,j!]$, let
$x\_n:=1/(j-1)!$, with $x\_1:=1$, so that $x\_n\ge1/n$ for all natural $n$. Also, for any fixed natural $k\ge2$ and all natural $j\ge k$, we have $kj!\in(j!,(j+1)!]$ and hence
$$\frac{x\_{kj!}}{x\_{j!}}=\frac{(j-1)!}{j!}\to0$$
as $j\to\infty$, so that
$$\inf\_{n\ge1}\frac{x\_{kn}}{x\_n}=0.$$


