The set of limit points of the sequence $1,\frac12,\frac14,\frac34,\frac18,\frac38,\frac58,\frac78,\frac1{16},\frac3{16},\ldots$

Solution 1:

The given sequence is a listing of all the rational numbers in $[0,1]$ of the form $\frac{m}{2^n}$ for some integers $m>0$ and $n\geq 0$. The set of rational numbers of this form turns out to be dense in $[0,1]$, so if by "limit points" you mean "points having infinitely many terms of the sequence in every neighborhood around them", then the answer is $[0,1]$.

Solution 2:

Let $x\in[0,1]$. For any $\epsilon>0$, there is a positive integer $n$ such that $\frac{1}{2^n}<\epsilon$. Then there is a positive integer $m$ such that $x\neq\frac{m}{2^n}\in (x-\epsilon,x+\epsilon)\cap[0,1]$. But $\frac{m}{2^n}$ is an element of the given sequence. Hence the set of limit points is $[0,1]$.

Solution 3:

Hint: Consider the first 8 terms (reordered): $$\frac{1}{8}, \frac{2}{8}, \frac{3}{8}, \frac{4}{8}, \frac{5}{8}, \frac{6}{8}, \frac{7}{8}, \frac{8}{8}$$

This pattern continues as the denominator grows throughout the sequence.