# $(a_n)$ is a convergent sequence in $\mathbb{R}$ with infinitely many positive and negative elements. Show that $(a_n)$ is a nullsequence. [duplicate]

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?

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

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 $$L=0$$
• 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 $L=0$. Apr 16, 2020 at 17:14