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Let $\{a_n\}$, a bounded sequence such that $a_n>0 \forall n>0$. Let:
$$\mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}{a_n} = 1$$
Prove that $\displaystyle\limsup_{n\to\infty} a_n\ge 1$.

My attempt:
It is given that $\mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}{a_n} = 1$.
Therefore, if the sequence is $a_n=1 \forall n>0$ then we've finished.
Otherwise, one term must be greater then one, while the other smaller then one.

Hence, There are two sub-sequences; One bounded from above and converges to $1$, and the other bounded bellow and converges to $1$ (because we know $a_n$ is bounded).

In any case, $\displaystyle\limsup_{n\to\infty}a_n=1$ as requested.

I think my proof is a bit handwaving. How can I "polish" it (Making it more rigorous)? I think I need to explain why there must be two sequences such that. Can you help me with that? Thanks!

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  • $\begingroup$ Your "one term must be greater then one, while the other smaller then one" is a non-sequitur. Consider for example $a_n=1+\frac1n$. $\endgroup$ – Hagen von Eitzen Jan 26 '14 at 15:22
  • $\begingroup$ If you are trying to prove that $a_n=1$ for some $n>N$, I don't think that would help. $\endgroup$ – jdoicj Jan 26 '14 at 15:22
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Hint: Suppose $\lim \sup a_n< 1$, then $a_n<1$ for very large $n$.

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  • 1
    $\begingroup$ I got it. and then, there's no way that $\mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}{a_n} = 1$. Right? A contradiction. $\endgroup$ – SuperStamp Jan 26 '14 at 15:27

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