Hi, I need help on the following:
Let $\displaystyle A = \{\frac{m}{2^n} \}$ where m,n are integers. $\displaystyle \frac{1}{3} $ is given as a limit point of A. Find a sequence in A that converges to $\displaystyle \frac{1}{3} $
Thanks
Hi, I need help on the following:
Let $\displaystyle A = \{\frac{m}{2^n} \}$ where m,n are integers. $\displaystyle \frac{1}{3} $ is given as a limit point of A. Find a sequence in A that converges to $\displaystyle \frac{1}{3} $
Thanks
What about $\displaystyle a_n := \frac{m}{2^n}$, where $\displaystyle m := \lceil \frac{2^n}{3}\rceil$?
Because then you get that $\displaystyle a_n \in A $, I think, and that
$\displaystyle \frac{1}{3}\leftarrow \frac{\frac{2^n}{3}-1}{2^n} \leq a_n = \frac{\lceil \frac{2^n}{3}\rceil}{2^n} \leq \frac{\frac{2^n}{3}+1}{2^n}\rightarrow \frac{1}{3}$