# Thread: Sequence and limit point

1. ## Sequence and limit point

Hi, I need help on the following:

Let $A = \{\frac{m}{2^n} \}$ where m,n are integers. $\frac{1}{3}$ is given as a limit point of A. Find a sequence in A that converges to $\frac{1}{3}$

Thanks

2. Originally Posted by RB06
Hi, I need help on the following:

Let $A = \{\frac{m}{2^n} \}$ where m,n are integers. $\frac{1}{3}$ is given as a limit point of A. Find a sequence in A that converges to $\frac{1}{3}$

Thanks
What about $a_n := \frac{m}{2^n}$, where $m := \lceil \frac{2^n}{3}\rceil$?
Because then you get that $a_n \in A$, I think, and that

$\frac{1}{3}\leftarrow \frac{\frac{2^n}{3}-1}{2^n} \leq a_n = \frac{\lceil \frac{2^n}{3}\rceil}{2^n} \leq \frac{\frac{2^n}{3}+1}{2^n}\rightarrow \frac{1}{3}$