Monotone Sequence $\frac{n}{n+1}$ I have to proof this sequence :
$(x_n) := \frac{n}{n+1}$ for $n\in \Bbb N^+$ is a monotone sequence.
I saw some examples of people using the induction method, so I tried it, but I got stuck when I was trying to prove the $A(k+1)$ is true.
Here's what I have done so far
Let $A(n) := x_n \le x_{n+1}$
$A(n)$ is true for all $n\ge1$. Assume that $A(k):=x_{k}\ge x_{k+1}$ is true, then show that $A(k+1)$ is also true.
I don't know where I should go next. I've been trying to add both side, but can't find the desired end. Should I change the proving method? For some backgrounds, before this question I had to prove that $(x_n)$ is convergent and bounded.
 A: No need to use induction here. If $n\in\Bbb N$, then$$x_{n+1}-x_n=\frac{n+1}{n+2}-\frac n{n+1}=\frac{1}{(n+2)(n+1)}>0$$and therefore your sequence is strictly increasing.
A: No need for induction. $\frac 1 n$ is decreasing so $1+\frac 1 n$ is also decreasing. This implies that $\frac 1 {1+\frac 1 n}$ is increasing. But $\frac 1 {1+\frac 1 n}=\frac n {n+1}$
A: Note that for every $n$
$$
\frac{x_{n+1}}{x_n}=\frac{\frac{n+1}{n+2}}{\frac{n}{n+1}}=\frac{n^2+2n+1}{n^2+2n}>1
$$
Therefore $\frac{x_{n+1}}{x_n}>1$, so $x_{n+1}>x_n$
A: As others said, induction is not the preferred way. Let us try anyway.
Base case:
$$\frac{0}{0+1}\le\frac{1}{1+1}$$ is true.
Inductive step:
We have to show that
$$\frac n{n+1}\le\frac{n+1}{n+2}\implies\frac{n+1}{n+2}\le\frac{n+2}{n+3}$$ or equivalently, subtracting $1$,
$$-\frac 1{n+1}\le-\frac1{n+2}\implies-\frac1{n+2}\le-\frac1{n+3}.$$
Then taking the opposite of the inverse,
$$n+1\le n+2\implies n+2\le n+3$$ which amounts to the trivial truth
$$1\le2\implies 2\le 3.$$
[By the way, this proof also contains the non-inductive proof.]
A: $1/(n+1)$ decreases with $n$, so that
$$
 x_n = \frac{n}{n+1} = 1 - \frac{1}{n+1}
$$
increases with increasing $n$.
Another option is to use real analysis: $f(x) = \frac{x}{x+1}$ is increasing on $[0, \infty)$ because
$$
 f'(x) = \frac{1}{(x+1)^2} > 0
$$
