Is $0$ a limit point of the sequence $(a_n)_{n=0}^{\infty}$, where $a_n:=(-1)^{3n+1}10^{-(3n+1)}$?

Question

Consider the sequence $(a_n)_{n=0}^{\infty}$, where $a_n:=(-1)^{3n+1}10^{-(3n+1)}$. Is $0$ a limit point for that sequence?

I see that, this sequence is "wandering" around $0$ and is bounded by $1$, and that hints me that 0 is possibly a limit point, though I tried to prove that via the definition by assuming some arbitrary $ε>0$ and $N\ge 0$, and so all that is left to show is to find some $n\ge N$ such that $|a_n-0|<ε$, or equivalently $10^{-(3n+1)}<ε$. But, I'm left utterly stumped. Also, I think that the inequality $10^{-m}<\frac{1}{m}$ is important to know.

Any suggestions as to how I could further proceed with this?

"the inequality $10^{-m}<\frac{1}{m}$ is important to know": indeed, and sufficient to conclude. — Anne Bauval, Jul 06 '23 at 19:11
@Vaskara_GRek_O, it is a duplicate question, please remove it. — Angelo, Jul 06 '23 at 19:35
$0$ is not just a limit point, it is the l8mit of the sequence. — Thomas Andrews, Jul 06 '23 at 20:06

score 2 · Answer 1 · answered Jul 06 '23 at 19:22

You've almost completed it. So continuing where you left of we can see: $$10^{-(3n+1)}<10^{-n}<\frac{1}{n}<\varepsilon$$

Hence we conclude that $\forall\varepsilon>0$ and $N\ge 0$, $\exists n\ge N, n\in \mathbb{N}$ such that $|(-1)^{3n+1}10^{-(3n+1)}-0|<\varepsilon$.

Is $0$ a limit point of the sequence $(a_n)_{n=0}^{\infty}$, where $a_n:=(-1)^{3n+1}10^{-(3n+1)}$?

1 Answers1