0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.