To prove $x_n<3$ for sequence $x_{n+1} = \frac{12(1+x_n)}{13+x_n}$ by induction

Question

Prove $x_n<3$ for a sequence given by $$x_{n+1} = \frac{12(1+x_{n})}{13+x_{n}}$$ where $x_1$ is positive real number less than $3$.

For $n = 1$ statement is trivial, but I am stuck at doing further. Thanks

Answer 1

Since $x_1<3$, assume that $x_n<3$ for some $n\in\mathbb{N}$.

Then $x_n<3\implies9x_n<27\implies 12+12x_n<39+3x_n\implies12(1+x_n)<3(13+x_n)$

$\;\;\;\implies\displaystyle x_{n+1}=\frac{12(1+x_n)}{13+x_n}<3$ $\;\;\;$(since $13+x_n>0$).

Answer 2

Since the base step is already given, we only need the induction step. Assume $x_n<3$. Then we need to prove that $x_{n+1}<3$. Notice that $\frac{12(1+x_n)}{13+x_n}$ is increasing on the given interval. If $x_n=3$ then

$$x_{n+1}=\frac{12(1+3)}{13+3}=3$$

Since $x_{n+1}$ increases with $x_n$, it will be less than $3$ when $x_n<3$, which is exactly our induction hypothesis.

Answer 3

$$\frac{12(1+x_n)}{13+x_n}=\frac{12(13+x_n)}{13+x_n}-\frac{12\cdot 12}{13+x_n} = 12 - \frac{12\cdot 13}{13+x_n} < 12 - \frac{12\cdot 12}{16} = 12-9=3$$ To see the inequality, notice that $x\mapsto \frac1{13+x}$ is decreasing, $x\mapsto -\frac1{13+x}$ is increasing. (For $x>-13$.)