# Convergence of a Sequence of Strictly Positive Real Numbers

(Disclaimer: This is a homework question.)

$$Let \, X = \, (x_{k})_{k\epsilon\mathbb{N}} \: be\,a\,sequence\,of\,strictly\,positive\,real\,numbers\,with\\ \lim_{k \rightarrow \infty } \tfrac{x_{k}}{x_{k+1}} > 1\\ prove\,that\,X\,is\,convergent\,and\,calculate\,lim(X).$$

Looking at the problem, my first thought was that $lim(X) = 0$.
Here's my attempt:
$$\tfrac{x_{k}}{x_{k+1}}\ convergent \implies \forall\,\varepsilon>0\,\,\exists\, k_{\varepsilon} \epsilon\,\mathbb{N}\,\, s.t.\,k\geq k_{\varepsilon} \implies |\tfrac{x_{k}}{x_{k+1}}-L| < \varepsilon\\with\, L>1.$$ The goal is to show that $\exists\, k_{\varepsilon} \epsilon\,\mathbb{N}\,\, s.t.\,k\geq k_{\varepsilon} \implies |x_{k}-c| < \varepsilon$ for some $c\,\epsilon\, \mathbb{R}$. Then this implies $lim(X) = c$. My guess is that $c=0.$ $$|x_{k} - L(x_{k+1})| < \varepsilon|x_{k+1}|\\|x_{k}| < (|L| + \varepsilon)|x_{k+1}|$$ This is the point where I get lost. Any hints or advice would be helpful.

The condition implies that there exist $k_0\in\mathbb{N}$ and $\epsilon>0$ such that for all $k>k_0$ $$x_{k+1}<\frac{x_{k}}{1+\epsilon}.$$ This implies $$x_{k+2}<\frac{x_{k+1}}{1+\epsilon}<\frac{x_{k}}{(1+\epsilon)^2}$$ $$...$$ $$x_{k+n}<\frac{x_{k}}{(1+\epsilon)^n}$$ Taking the limit as $n\to\infty$ you get $\lim_{k\to \infty} x_k=0.$