Show that a sequence is bounded, monotone, using The Convergence Theorem

[cbarker1]

Dear Every one,

In my book, Basic Analysis by Jiri Lebel, the exercise states
"show that the sequence $\left\{(n+1)/n\right\}$ is monotone, bounded, and use the monotone convergence theorem to find the limit"

My Work:
The Proof:
Bound
The sequence is bounded by 0.
$\left|{(n+1)/n}\right| \ge 0$
$n+1\ge0\, \forall\, n \in \Bbb{N}$

Monotone
$\frac{n+1}{n} \ge 0$$\frac{\left(n+1\right)+1}{n+1}\le\frac{n+1}{n}$
$\frac{n+2}{n+1}\le\frac{n+1}{n}$
$n^2+2n\le n^2+2n+1$
the sequence is increasing monotone.
Applying the theorem.
$\lim_{{n}\to{\infty}} \frac{n+1}{n}=\sup{{\frac{n+1}{n}:n\in\Bbb{N}}}$
Let A=$\sup{{\frac{n+1}{n}:n\in\Bbb{N}}}$
We know the sup A is less than or equal to 1 as 1 is an upper bound. Take a number $$b\le1$$ such that $$b\ge \frac{n+1}{n}$$.
$$bn\ge n+1$$
$$bn-n\ge1$$
$$n(b-1)\ge1$$
By the Archimedean Property..."

What need I to do in order to finish the proof?

THanks
CBarker1

 

[S.G. Janssens]

the sequence is increasing monotone.

I would say it is decreasing.

We know the sup A is less than or equal to 1 as 1 is an upper bound.

Are you sure $1$ is an upper bound?

Take a number $$b\le1$$ such that $$b\ge \frac{n+1}{n}$$.
$$bn\ge n+1$$
$$bn-n\ge1$$
$$n(b-1)\ge1$$
By the Archimedean Property..."

What need I to do in order to finish the proof?

The sequence has a lower bound. Indeed, as you wrote, zero is a trivial lower bound. One is a slightly less trivial lower bound. Can you prove this?

The sequence is also monotonously decreasing. So, you know it has a limit in $\mathbb{R}$. Now you could try to show that the infimum of the sequence equals one, from which the value of the limit follows.