Prove that the least upper bound of the sequence $\left\{ \frac{2^n - 1}{2^{n-1}} \right\}$ is $2$ It can be easily verified that the sequence $\left\{ \dfrac{2^n -1} {2^{n-1}} \right\}$ is strictly positive, the task now is to prove that 2 is an upper bound.
Proof:
Suppose for the sake of contradiction that 2 is not the least upper bound of the sequence. Let $x_n$ be an element of the sequence. Using the density of the real numbers in combination with the fact that the sequence is strictly increasing we can write the following:
$$ x_i < 2 < x_j , \quad j>i  $$
That is, we can find a $j \in \mathbb{N}$ such that:
\begin{align}
0 &< x_j - 2 \\
  &<\dfrac{2^j - 1}{2^{j-1}} - 2 \\ 
&< \dfrac{2^j - 1 - 2(2^{j-1})}{2^{j-1}} \\
&< (-1)\dfrac{1}{2^{j-1}} \\ 
\end{align}
So we arrive at a contradiction.
I just want to check to see if my approach makes sense to this problem.
 A: $$\frac{2^n-1}{2^{n-1}}<2$$
multiply both sides by $2^{n-1}$
$$2^n-1<2^n$$
which is true for any $n$.
Furthermore $$\frac{2^n-1}{2^{n-1}}=\frac{2^n}{2^{n-1}}-\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}$$
For any $\varepsilon >0$ there exists $N$ such that if $n>N$ and $n\in\mathbb {N}$ then
$$\frac{1}{2^{n-1}}<\varepsilon$$
we take $N=1-\log _2 \varepsilon $.
Therefore $2$ is the least upper bound of the sequence $\{\frac{2^n-1}{2^{n-1}}\}$
A: Your reasoning is incomplete, because it only establishes that there is no $j$ such that $x_j > 2$.  It does not demonstrate that $2$ is the least such bound; e.g., if you replace $2$ with $4$, your reasoning would still apply.  But $4$ is not the least upper bound.
In order to show that $2$ is the least upper bound, one way is to establish that if $B$ is the set of all upper bounds for $x_n$, that $2$ is the smallest member of $B$.
A: Your solution is fine, you could also notice that $\frac{2^n-1}{2^{n-1}}=\frac{2^n}{2^{n-1}}-\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}}\leq2$ as $\frac{1}{2^{n-1}}\geq0$
A: Your answer is okay but it's ludicrously complicated.
Why not just say $\frac {2^n -1}{2^{n-1}} < \frac {2^n}{2^{n-1}} = 2$?
That's it.  You are done.
But is that all of the question?
Are you sure you weren't also being as to prove that $2$ is the least upper bound.
In that case if $x < 2$ you must prove there exists a $x < \frac {2^n-1}{2^{n-1} } < 2$ no matter how close $x$ is to $2$.
$x < \frac {2^n-1}{2^{n-1} } = \frac {2^n}{2^{n-1}} - \frac 1{2^{n-1}} = 2-\frac 1{2^{n-1}} < 2 \iff$
$x-2 <  -\frac 1{2^{n-1}} < 0\iff$
$0 < \frac 1{2^{n-1}} < 2-x \iff$
$x < 2$ and $2^{n-1} > \frac 1{2-x}$.
If we know that $2^m$ is unbounded then we know that there is an $m = n-1$ so that $2^{n-1} > \frac 1{2-x} $ for any $x < 2$.
If $\{2^m\}$ were bounded above then $\sup 2^m$ would exist.  And $\sup 2^m -1$ wouldn't be an upper bound so there would be a $k$ so that $\sup 2^m -1 < 2^k < \sup 2^m$  So $2\sup 2^m - 2 < 2^{k+1}$.  But as $\sup 2^m  > 2=2^2$ as must have $2\sup 2^m - 2 > \sup 2^m$ and so $\sup 2^m < 2^{k+1}$ which is a contradiction.
