1
$\begingroup$

The definition for a null sequence is following:

$\forall \epsilon > 0, \exists n_0 \in \mathbb{N}, \forall n \geq n_0: |a_n -0|< \epsilon $

We only know that the sequence is convergent and that there are infinitely many positive elements. How can I prove that $a_n$ is a null-sequence with my definition?

$\endgroup$
3
  • $\begingroup$ You know that the sequence is convergent. Now assume that the limit is not zero... $\endgroup$
    – Martin R
    Apr 16, 2020 at 15:53
  • 4
    $\begingroup$ Does this answer your question? show that if ${a_n} \to a$, and $a>0$, then $\exists N$ such that $a_n>0$ for $n \ge N$ $\endgroup$
    – Martin R
    Apr 16, 2020 at 15:54
  • 3
    $\begingroup$ Every subsequence converges to the same limit, thus pick two subsequences, one consisting of entirly positive elements and the other of fully negative elements $\endgroup$
    – user732848
    Apr 16, 2020 at 15:54

1 Answer 1

4
$\begingroup$

Since this sequence $(x_n)$ is convergent, hence every subsequence of this sequence converges to the same limit $L$ (Which you assume to be non-zero, because if it were zero you would be done at that point), and thus pick two subsequences $({x_n}_i), ({x_n}_j)$ such that $({x_n}_i)$ consisting entirely of positive elements of $(x_n)$ and $({x_n}_j)$ wholly consists of negatives (this is possible as there are infinitely many such). $({x_n}_i)$ converges to the limit $L$ as each term is ${x_n}_i >0$ while the second subsequence $({x_n}_j)$ converges to $L$ but as each term is ${x_n}_j<0$ we should have $L=-L$ or $L=0$

$\endgroup$
7
  • $\begingroup$ Why do we assume the limit to be non-zero? It's a null-sequence? $\endgroup$ Apr 16, 2020 at 16:13
  • 2
    $\begingroup$ If the limit is zero then you are already done! $\endgroup$
    – user732848
    Apr 16, 2020 at 16:15
  • $\begingroup$ sorry for my confusion. The solution with showing the Limit of the subsequences is great and thank you very much for your help. I still don't really the the full picture though. Could you please write me a few more words if you have the time. Thank you very much sir. $\endgroup$ Apr 16, 2020 at 16:19
  • 1
    $\begingroup$ @PeterSzilas yes you are right i will detail my solution soon. $\endgroup$
    – user732848
    Apr 16, 2020 at 16:25
  • 2
    $\begingroup$ Shamim. Perhaps this is better now:. $a_{n_k}$ positive subsequence converges to $L\ge 0$. Similarly the neg subsequence $a_{n_l}$ converges to $L\le 0$.Hence $L=0$. $\endgroup$ Apr 16, 2020 at 17:14

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