Let $(x_n)_{n \in \mathbb{N}}$ be the sequence defined by $x_1=1$ and $ x_{n+1}=\dfrac{4+3x_n}{3+2x_n} \text{for} \hspace{0.2 cm }n=1,2,3…$
Show that $(x_n)_{n \in \mathbb{N}}$ is an increasing sequence
First I used to prove $x_{n}\leq x_{n+1}$ by using mathematical induction but it is not success then I used a function
$f(x)=\dfrac{4+3x}{3+2x} $ which first derivative is always positive thus I conclude that $f(x)$ is an increasing function thus $x_{n}\leq x_{n+1}$
Can anyone help me I don't know what I am doing is correct or not?
If anyone help me to prove this using mathematical induction it is better!!
Best Answer
You have differentiated the function and shown that it is positive, you can also check that $x_1<x_2$ manually. Then you have from the mean value theorem
$$x_{n+1}-x_n=\frac{4+3x_n}{3+2x_n}-\frac{4+3x_{n-1}}{3+2x_{n-1}}=f'(c)\cdot(x_n-x_{n-1})>0, c \in (x_{n-1},x_n)$$
where we used the induction hypothesis $x_n>x_{n-1}$ and the fact that the derivative is bigger than zero.
Only induction:
First use induction to show that $x_n$ is positive, this is very easy.
Then we get:
Check first manually that $x_1<x_2$. Assume now that $x_{n-1}<x_n$ we get
$$x_{n+1}-x_n=\frac{4+3x_n}{3+2x_n}-\frac{4+3x_{n-1}}{3+2x_{n-1}}\\=\frac{(4+3x_n)(3+2x_{n-1})-(4+3x_{n-1})(3+2x_n)}{(3+2x_n)(3+2x_{n-1})}\\=\frac{12+8x_{n-1}+9x_n+6x_nx_{n-1}-12-8x_n-9x_{n-1}-6x_nx_{n-1}}{(3+2x_n)(3+2x_{n-1})}\\=\frac{x_n-x_{n-1}}{(3+2x_n)(3+2x_{n-1})}>0,$$
where we have used the induction hypothesis, $x_n>x_{n-1}$ and the fact that $x_n,x_{n-1}$ is positive.