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Show that the sequence $(x_n)$ with $x_1 = 1$ and $x_{n+1} = \dfrac{2x_n + 1}{x_n+3}$ for $n ∈ \Bbb{N}$ converges and determine its limit.

I understand how to show the limit of sequence, but how could I show that it is decreasing? I tried to do it by induction: $$ x_{n+1} =\frac{2x_n + 1}{x_n + 3} \\ x_{n+2} = \frac{2x_{n+1} + 1}{x_{n+1} + 3} \\ x_{n+2} < x_{n+1} \\ \frac{2x_{n+1} + 1}{x_{n+1} + 3} < \frac{2x_n + 1}{x_n + 3} \\ \text{Then I put } x_{n+1} =\frac{2x_n + 1}{x_n + 3} \text{ at the left part of the equation} \\ \frac{2{(\frac{2x_n + 1}{x_n + 3} )} + 1}{(\frac{2x_n + 1}{x_n + 3}) + 3} < \frac{2x_n + 1}{x_n + 3} \\ \frac{x_n^2 + 4x_n + 3}{x_n^2 + 5x_n + 6}< \frac{2x_n + 1}{x_n + 3} \\ ({2x_n +1})({x_n^2 + 5x_n + 6})< ({x_n^2 +4x_n +3})({x_n + 3}) \\ {x_n^3+4x_n^2+2x_n-3>0} $$ And I don't know what to do next and I can't determine its limit until I proof that it's converges

I don't need a name
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    Use induction. And please use Latex formatting. – GReyes Oct 25 '23 at 01:44
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    I tried to use induction, but I did something wrong( – I don't need a name Oct 25 '23 at 01:46
  • Can you show your attempt? – GReyes Oct 25 '23 at 01:57
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    To show it's decreasing, find $x_{n+1}-x_n$. You should get a quadratic that is negative for all $x_n>\frac{\sqrt 5-1}2=0.618...$, which proves that $x_{n+1}<x_n$ – Phobo Havuz Oct 25 '23 at 02:01
  • $$ x_{n+1} =\frac{2x_n + 1}{x_n + 3} \ x_{n+2} = \frac{2x_{n+1} + 1}{x_{n+1} + 3} \ x_{n+2} < x_{n+1} \ \frac{2x_{n+1} + 1}{x_{n+1} + 3} < \frac{2x_n + 1}{x_n + 3} \ \text{Then I put } x_{n+1} =\frac{2x_n + 1}{x_n + 3} \text{ at the left part of the equation} \ \frac{2{(\frac{2x_n + 1}{x_n + 3} )} + 1}{(\frac{2x_n + 1}{x_n + 3}) + 3} < \frac{2x_n + 1}{x_n + 3} \ \frac{x_n^2 + 4x_n + 3}{x_n^2 + 5x_n + 6}< \frac{2x_n + 1}{x_n + 3} \ ({2x_n +1})({x_n^2 + 5x_n + 6})< ({x_n^2 +4x_n +3})({x_n + 3}) \ {x_n^3+4x_n^2+2x_n-3>0} $$ And I don't know what to do next( – I don't need a name Oct 25 '23 at 02:46
  • https://math.stackexchange.com/questions/3885418/finding-the-closed-form-of-a-recursive-sequence/3885693#3885693 – Claude Leibovici Oct 25 '23 at 04:02
  • Calling $\phi(x) = \frac{2x+1}{x+3}$ we have $|\phi'(x)| < 1$ for $x\ge 1$ then follows $|x_{k+1}-x_k| \ge |x_k-x_{k-1}|$ – Cesareo Oct 25 '23 at 08:43

2 Answers2

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Note that:

$$x_{n+1}=2-\frac{5}{x_n+3}$$

Since $x_1=1$ and $x_2=0.75$, so $x_2<x_1$. Suppose that for some $n\in \mathbb{Z}^+$ we have that $x_{n+1}<x_{n}$. Then:

$$x_{n+1}+3<x_{n}+3$$ $$\frac{1}{x_{n}+3}<\frac{1}{x_{n+1}+3}$$ $$2-\frac{1}{x_{n+1}+3}<2-\frac{1}{x_{n}+3}$$ $$x_{n+2}<x_{n+1}$$

So by first principle of mathematical induction the it follows that $x_{n+1}<x_{n}$ for all $n\in \mathbb{Z}^+$. Now all that remains to be shown is that this is bounded below then by Monotone Convergence Theorem you can easily deduce the convergence and therefore find the limit.

Shyam Tripathi
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Given that $x_1=1$ and $x_{n+1}=\frac{2 x_n+1}{x_n+3} \cdots (*)$ for any integer $n\ge 1.$ We note that $x_n$ is positive and hence bounded below by $0$.

Let the proposition be $P(n): x_{n+1}\le x_{n}$, where $n\in \mathbb{N}.$ $x_2=\frac34<x_1 \Rightarrow P(1) $ is true.

Assume that $P(k)$ is true for some natural number $k$, then

$$ \begin{aligned} x_{k+2} & =\frac{2 x_{k+1}+1}{x_{k+1}+3} \\ & =2-\frac{5}{x_{k+1}+3} \\ & <2-\frac{5}{x_{k+3}} \\ & =\frac{2 x_k+1}{x_k+3} \\ & =x_{k+1} \end{aligned} $$ Therefore $P(k+1)$ is also true. By M.I., $P(n)$ is always true for all $n\in \mathbb{N}.$ So the sequence $x_n$ is decreasing and bounded below and hence is convergent, say, to $L$. Taking limit of $n$ to infinity on both sides of $(*)$ yields

$$ \begin{aligned} & L=\frac{2 L+1}{L+3} \\ \Leftrightarrow \quad & L^2+L-1=0 \\ \Leftrightarrow \quad & L=\frac{-1+\sqrt{5}}{2} \end{aligned} $$

Lai
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