Proving $\frac{n}{n+1} < \frac{n+1}{n+2}$ by induction?

Question

I have the inequality

$\frac{n}{n+1} < \frac{n+1}{n+2}$

I'm not sure how to go about proving it. I've started by testing with n = 1, which results in

$\frac{1}{2} < \frac{2}{3}$ which is true

I then assume true for n = k and have to prove that it is true for n = k + 1, but I don't know how to start manipulating

$\frac{k}{k+1} < \frac{k+1}{k+2}$

to become

$\frac{(k+1)}{(k+1)+1} < \frac{(k+1)+1}{(k+1)+2}$

How do I go about doing this?

Answer 1

You don't need induction, since it is enough to show $$n(n+2)<(n+1)^2$$

Answer 2

This only requires basic inequation manipulation ($n \in \mathbb{N}$):

$${n \over n+ 1} < {n + 1\over n+ 2}$$

$$\iff n(n+2) < (n + 1)^2$$

$$\iff n^2+2n < n^2 + 2n + 1$$

$$\iff 0 < 1$$

Done.

Answer 3

Use : if $ 0 < a < b \Rightarrow \dfrac{a}{b} < \dfrac{a+1}{b+1} $, with $a = k, b = k+1$, and once more $ a = k+1, b = k+2$.

Answer 4

It is a well-known inequality that, if $a,b,h>0$ then $$\begin{cases} \dfrac ab<1\implies \dfrac ab<\dfrac{a+h}{b+h}<1,\\ \dfrac ab>1\implies \dfrac ab>\dfrac{a+h}{b+h}>1. \end{cases}$$

Answer 5

if you need induction:

\begin{gather*} n(n+2)<(n+1)^2 \\ \begin{aligned} (n+1)(n+3)={}&n(n+2+1)+(n+3)=(n(n+2)+n+n+3)<(n+1)^2+2n+3={} \\ {}={}&(n^2+4n+4)=(n+2)^2. \end{aligned} \end{gather*}

Answer 6

Here is another way: Let $f(x) = { x \over x+1}$ and show that $f'(x) >0 $ for all $x \ge 0$. Then $f$ is strictly increasing and so $f(n) < f(n+1)$ for all $n \ge 0$.

Answer 7

Note that $$ \frac{n}{n+1} = 1 - \frac{1}{n+1},$$ $$ \frac{n+1}{n+2} = 1 - \frac{1}{n+2},$$ so actually you just need to show that $$\frac{1}{n+1} > \frac{1}{n+2}.$$