Condition for the sequence $\left\{x_n\right\}$ being monotone where $x_1>0$ and $x_{n+1}=\frac{3(1+x_n)}{5+x_n}$

Question

A sequence $\left\{x_n\right\}$ is given by $x_1>0$ and $x_{n+1}=\frac{3(1+x_n)}{5+x_n}$ for all $n\in\mathbb{N}$.

How can I prove that

1. The sequence $\left\{x_n\right\}$ is monotone increasing if $0<x_1<1$.
2. The sequence $\left\{x_n\right\}$ is monotone decreasing if $x_1>1$.

I cannot prove this. Should I prove this by assuming first that the sequence is monotone increasing and thereby deduce the range of $x_1$, or what?

Please anyone help me solving it. Thanks in advance.

Answer 1

For the case $0<x_1<1$, we'll prove by induction that $0<x_n<1$.

Induction basis: $0<x_1<1$.

Induction step: Assume $0<x_n<1$. The inequality $0<x_{n+1}$ is obvious from the recursive formula because of $0<x_n$. Additionally, we have

$$x_{n+1}=\frac{3(1+x_n)}{5+x_n}=\frac{3+3x_n}{5+x_n}<\frac{5+x_n}{5+x_n}=1.$$

Thus, the induction is complete and $0<x_n<1$ holds for all $n$.

This implies

$$x_{n+1}=\frac{3(1+x_n)}{5+x_n}>\frac{3(1+x_n)}{5+1}=\frac{1}{2}(1+x_n)>\frac{1}{2}(x_n+x_n)=x_n$$

and hence, the sequence is monotonically increasing.

The second statement for the case $x_1>1$ is proved analogously.

Answer 2

$$\frac{x_{n+1}-1}{x_{n+1}+3}=\frac{\frac{3(1+x_n)}{5+x_n}-1}{\frac{3(1+x_n)}{5+x_n}+3}=\frac{2x_n-2}{6x_n+18}=\frac{1}{3}\cdot\frac{x_n-1}{x_n+3}.$$ Thus, $$\frac{x_n-1}{x_n+3}=\frac{x_1-1}{x_1+3}\cdot\left(\frac{1}{3}\right)^{n-1}$$ or $$1-\frac{4}{x_n+3}=\frac{x_1-1}{x_1+3}\cdot\left(\frac{1}{3}\right)^{n-1}.$$ Now, we see that for $0<x_1<1$ the sequence $x$ increases (because the expression $\frac{x_1-1}{x_1+3}\cdot\left(\frac{1}{3}\right)^{n-1}$ increases) and for $x_1>1$ it decreases.