Lower bound for $\frac{x_{kn}}{x_n}$, where $(x_n)_{n\in\mathbb{N}}$ is a non-increasing sequence in [0,1] with $x_n\ge\frac{1}{n}$

Question

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 $\frac{x_{kn}}{x_n}\ge\frac{1}{k}$.

Answer 1

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 $$\inf_{n\ge1}\frac{x_{kn}}{x_n}=0.$$