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#### OhMyMarkov

##### Member

- Mar 5, 2012

- 83

I am told that the limit of $\frac{x_{n+1}}{x_n}$ is $L>1$. I am asked to show that $\{x_n\}$ is not bounded and hence not convergent.

This is what I got so far:

Fix $\epsilon > 0$, $\exists n_0 \in N$ s.t. $\forall n > n_0$, we have

$|\frac{x_{n+1}}{x_n}-L|<\epsilon$.

Rearranging terms, we have:

$(L-\epsilon)\cdot x_n<x_{n+1}$

I'm stuck here, I want to show that there is an $M$ s.t. $x_n>M \; \forall n>n_1$ for some $n_1$. Thus the sequence is unbounded and converges to infinity. What I am thinking about is the following:

(1) Show that the sequence is increasing, then I get $x_n$ > $x_{n_1}$ which is a constant, but I don't know how.

(2) ... but then I don't know what to do with $\epsilon$ because it could be potentially bigger than $L$.