If $\lim_{x \to \infty} x_n = \liminf_{n \to \infty} x_n = \limsup_{n \to \infty} x_n= -\infty$ does it exist a convergent subsequence of $x_n$? I've learned that the limit of a convergent subsequence is called accumulation point,$\limsup$ as the greatest accumulation point and $\liminf$ as the smallest accumulation point. Then if $\limsup$ and $\liminf$ are $-\infty$ then all limits of all subsequences of $x_n$ are between $-\infty$ and $-\infty$ i.e. any of the sequences converges. I'm not really sure if this argument is correct. Could you help me please?

  • $\begingroup$ If $\{x_n\}$ converges (including to the general points $\pm \infty$), then all its convergent subsequences must converge to the same limit. $\endgroup$ – Zhanxiong Dec 6 '15 at 19:52
  • $\begingroup$ @Solitary could you please say where can I find the theorem you're using to say this? I'm reading "Elementary Analysis" of Ross and there is mentioned that this property is only applicable to convergent sequences. $\endgroup$ – Jelly Belly Dec 6 '15 at 19:58
  • $\begingroup$ I give you a simple proof in the following answer, hope it helps. $\endgroup$ – Zhanxiong Dec 6 '15 at 23:12

Suppose $\{x_n\}$ diverges to $+\infty$ but one subsequence $\{x_{n_k}\}$ converge to $-\infty < L < +\infty$.

Since $\lim_{n \to \infty} x_n = +\infty$, there exists $N \in \mathbb{N}$ such that

$$x_n > L + 1 \text{ for all } n > N. \tag{1} $$

On the other hand, since $\lim_{k \to \infty} x_{n_k} = L$, there exists $K \in \mathbb{N}$ such that

$$|x_{n_k} - L| < 1 \text{ for all } k > K. \tag{2}$$

Now for any $k$ sufficiently large such that $k > K$ and $n_k > N$, by $(2)$ we have $$x_{n_k} = L + x_{n_k} - L < L + 1,$$ which is contradict with $(1)$.

Use the similar argument, you can cover other cases.


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