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I have $\lim_{x_n\to\infty}=0$ and $\lim_{x_n\to\infty}x_{n+1}/x_n=L$. How do I show that $|L|<1$.

I tried to break it up and show that $L>-1$ and $L<1$, I let $\epsilon=|L|/2, \text{ and}\, \epsilon=|1-L|/2 $ but it didn't work out too well. I have too many case to work with.

Can anyone give me a better idea please?

1 Answers 1

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First assume everything is positive-the other cases are similar. If $L \gt 1$, there must be some $N,\epsilon$ such that if $n \gt N,\frac {x_{n+1}}{x_n} \gt 1+\epsilon$. Then the series is bounded below by $x_N(1+\epsilon)^{n-N}$ which increases without bound, contradicting the $\lim_{x_n \to +\infty}=0$