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.


closed as off-topic by rtybase, cmk, Shogun, Xander Henderson, José Carlos Santos Jun 12 at 20:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – rtybase, cmk, Shogun, Xander Henderson, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ You neediof $0<x<1$ then $x<\frac{3(1+x)}{5+x}<1.$ And likewise, if $x>1$ then $1<\frac{3(1+x)}{5+x}<x.$ $\endgroup$ – Thomas Andrews Jun 11 at 20:51

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


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

This implies


and hence, the sequence is monotonically increasing.

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


$$\frac{x_{n+1}-1}{x_{n+1}+3}=\frac{\frac{3(1+x_n)}{5+x_n}-1}{\frac{3(1+x_n)}{5+x_n}+3}=\frac{2x_n-2}{6x_n+18}=\frac{1}{3}\cdot\frac{x_n-1}{x_n+3}.$$ 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.

  • $\begingroup$ Can the down-voter explain, why did you do it? $\endgroup$ – Michael Rozenberg Jun 12 at 7:07
  • $\begingroup$ I didn't downvote, but I was downvoted too ... It looks like a revenge for the what looks like a downvoted question. I voted to close it, it lacks of details anyway. $\endgroup$ – rtybase Jun 12 at 7:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.