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Let $\{a_n\}$ be a sequence of non-zero real numbers. Then, can we say that it has a subsequence $\{a_{n_k}\}$ such that $\lim\frac{a_{n_{k+1}}}{a_{n_k}}$ exists and belongs to the set $\{0,1,\infty\}$?

I think yes, because, if the sequence were convergent, then, the values $0,1$ can easily be found. But the doubt comes when the sequence were divergent and value of limit is $\infty$. Any ideas. Thanks beforehand.

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    $\begingroup$ The result is true. Three cases. 1. If $(a_n)$ and $(a_n^{-1})$ are bounded, then a subsequence converges to $\ell\ne0$ hence the ratios of this subsequence converge to $1$. 2. If $(a_n)$ is unbounded, a subsequence $(a_{\phi(n)})$ diverges to $+\infty$, say, hence the ratios of a subsubsequence converge to $+\infty$. 2'. Idem when a subsequence $(a_{\phi(n)})$ diverges to $-\infty$. 3. Finally, the same holds when $(a_n^{-1})$ is unbounded, but with the limit $0$. $\endgroup$ – Did May 8 '17 at 19:05
  • $\begingroup$ Hint: if $a_n$ is unbounded, you're done. (Why?) If $a_n$ is bounded for all sufficiently large $n$, it must have an accumulation point within those bounds. (Why?) If that accumulation point is non-zero, you're done. If the accumulation point is zero, then you should be able to find a version of the first argument... $\endgroup$ – Steven Stadnicki May 8 '17 at 19:06
  • $\begingroup$ @StevenStadnicki how come if $\{a_n\}$ is bounded for large $n$, a limit point should exist? It is not given monotonic, right? $\endgroup$ – vidyarthi May 8 '17 at 19:12
  • $\begingroup$ Lookup the Bolzano–Weierstrass theorem to get an understanding of that. $\endgroup$ – David May 8 '17 at 19:19
  • $\begingroup$ @vidyarthi Be careful: a limit point doesn't necessarily exist, but an accumulation point will. (An accumulation point is essentially a limit point of a subsequence.) The simplest intuition might be through the infinite pigeonhole theorem (i.e. Konig's Lemma) : split the interval of your bounds into two subintervals. Then the sequence must contain infinitely many members in one or the other of the subintervals (or possibly both); choose a subinterval which has infinitely many sequence members, create the subsequence of all sequence members within that subinterval, and recurse. $\endgroup$ – Steven Stadnicki May 8 '17 at 21:16

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