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Let $\{a_n\}$ be a sequence of non-zero real numbers.

Show that it has a subsequence $\{a_{n_k}\}$ such that $\lim \dfrac{a_{n_{k+1}}}{ {a_{n_k}}}$ exists and belongs to $\{0,1,\infty\}$.

I am finding the above problem false. If I take $(a_n)_n=(e^{-n})_n$ then any sub-sequence of $a_n$ is $e^{-n_k}$ but $\lim \dfrac{a_{n_{k+1}}}{ {a_{n_k}}}=\dfrac{e^{-n-1}}{e^{-n}}=\dfrac{1}{e}\notin \{0,1,\infty\}$.

Edits:By @Henry's comment I am sure the problem is true.But how should I find the sub-sequence.Please give some hints.

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    Hint: first show that every sequence with unequal terms has an strictly monotone subsequence – 2017-01-16

2 Answers 2

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Notice that $n_k+1 = n_{k+1}$ is false most of the times so your counterexample does not work. Take $n_k = 2^k$, then $$ \frac{a_{n_{k+1}}}{a_{n_k}} = \frac{e^{-2^{k+1}}}{e^{-2^k}} = \exp(-2^k) \to 0 $$

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If the sequence $(a_n)_{n\geq0}$ is unbounded then WLOG we may assume that for each $N\geq0$ there is an $n$ with $a_n>N$. In this case let $n_0$ be the first $n$ with $a_n>0$, and choose for any $k\geq1$ an $n_k>n_{k-1}$ with $a_{n_k}\geq k \>a_{n_{k-1}}$. It follows that $\lim_{k\to\infty}{a_{n_k}\over a_{n_{k-1}}}=\infty$.

If the sequence $(a_n)_{n\geq0}$ is bounded by Bolzano's theorem there is a subsequence $(a_{n_k})_{k\geq0}$ converging to some $\lambda\in{\mathbb R}$. If $\lambda\ne0$ then $\lim_{k\to\infty}{a_{n_k}\over a_{n_{k-1}}}=1$. If $\lambda=0$ then after sieving we may assume $\lim_{n\to\infty} a_n=0$, and a second sieving allows to assume that, e.g., $a_n>0$ for all $n\geq0$. In this case put $n_0=0$, and choose for any $k\geq1$ an $n_k$ with $0

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    Will you please explain the first paragraph.i.e how to choose $n_k$ with $a_{n_k}\ge ka_{n_{k-1}}$ – 2017-03-12
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    After $n_{k-1}$ has been chosen the large number $N:=k a_{n_{k-1}}$ is determined. Since the original sequence contains a subsequence converging to $\infty$ you can find an $n_k>n_{k-1}$ such that $a_{n_k}>N$. – 2017-03-12