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Let $a_n \neq 0$ and $\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right| = q,\ q > 1$. Prove that $\lim_{n \to \infty} a_n = 0$

It's easy to prove that if $x_n > 0$ and $\lim \frac{x_{n+1}}{x_n} < 1 \Rightarrow \lim x_n = 0$. Just use squeeze theorem and show $0 < x_n < x_n(\frac{1 + q}{2})^{(n - N)}$, where $q = \lim_{n \to \infty}\frac{x_{n + 1}}{x_n}$ and $n > N$.

However It's hard for me to apply squeeze theorem, when $x_n$ is not neccesarily positive.

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  • $\begingroup$ If you want to work with a sequence such that $a_n>0$, you can always pick a subsequence such that this is true. What if no such subsequence exists? That shouldn't be too hard either, given your approach. $\endgroup$ – Theoretical Economist Dec 17 '16 at 10:25
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Hint. Apply what you wrote

"It's easy to prove that if $x_n > 0$ and $\lim \frac{x_{n+1}}{x_n} < 1 \Rightarrow \lim x_n = 0$. Just use squeeze theorem and show $0 < x_n < x_n(\frac{1 + q}{2})^{(n - N)}$, where $q = \lim_{n \to \infty}\frac{x_{n + 1}}{x_n}$ and $n > N$."

to $|a_n|$.

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