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I have a problem with exercises with sequences given by recursion when I need to "prove convergence and find limit if it exists" and I am given a recursion of that kind:

$$ x_{1+n} = \frac{1}{2 + x_{n}}, x_1 \in (0 ; \infty)$$

It is fairly easy to find the limit - I just assume that the limit exists in $ \mathbb{R}$ and then use arithmetic properties of limits: $$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} x_{n}$$ $$\lim_{n \to \infty} x_{n} = l, l \in \mathbb{R}>0$$

Taking my recursion: $$l = \frac{1}{2 + l}$$ $$l^2 +2l - 1 = 0$$ $$l_1 = \sqrt{2} - 1 \in D$$ $$l_2 = -1 - \sqrt{2} \notin D$$

So my only possible limit in $ \mathbb{R}$ is $l = \sqrt{2} - 1$. That is if I can actualy prove that the limit exists - that is: the sequence is monotonous and bounded. And here is my problem - it is just impossible to analyse without computer the difference of:

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

In search of limits I just multiply both sides of equation by $ \lim_{n \to \infty} x_{n} = l$ and it is impossible to do so here, so I get: $$x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$$

Then I can't tell when it is bigger than $0$ to analyse monotonicity and I can't the see for which values o $n$ which values of $n+1$ i get (to get the boundary) because min value gets crazy.

So I just waneted to ask - am I missing something? Is it possible to make here $x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$ an equality with $0$ and analyse simpler function (red one on the picture)?

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theboyboy
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3 Answers3

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$$X_{n+1}=\frac{1}{2+X_n} \implies 2 X_{n+1}+X_{n+1}X_n=1$$ Let $X_n=\frac{Y_{n-1}}{Y_n}$, then $$2 \frac{Y_{n}}{Y_{n+1}}+\frac{Y_n}{Y_{n+1}}\frac{Y_{n-1}}{Y_n}=1 \implies 2Y_n+Y_{n-1}=Y_{n+1}.$$ Let $Y_n=t \implies t^2-2t-1=0 \implies t=1\pm \sqrt{2}.$ So $$Y_n=p(1+\sqrt{2})^n+q (1-\sqrt{2})^{n} $$ $$\implies X_n=\frac{(1+\sqrt{2})^{n-1}+r(1-\sqrt{2})^{n-1}}{(1+\sqrt{2})^{n}+r(1-\sqrt{2})^{n}}, r=q/p.$$ $$\lim_{n \to \infty}X_{\infty}=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$$

Z Ahmed
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  • Cool, but how do you know that the limit actually exist (the sequence is monotonous and bounded)? That is my problem. – theboyboy Nov 24 '20 at 06:40
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Although $x_1$ can be any positive number, all the terms starting from $x_2$ are less than $\frac 12$, so cannot be far from your limit. An approach that can be useful is to write one term as the limit plus an error term, so here let $x_i=\sqrt 2-1+\epsilon$ Then $$x_{i+1}=\frac 1{2+x_i}=\frac 1{1+\sqrt 2 + \epsilon}\\ x_{i+1}=\frac{\sqrt 2-1}{1+(\sqrt 2-1)\epsilon}\\ x_{i+1}\approx (\sqrt 2-1)-(\sqrt 2-1)^2\epsilon$$ where I have used the first order approximation to $\frac 1{1+\epsilon}$. We see from this that the error is decreased by a factor about $6$ every step, so the sequence will converge. To be more formal, you can bound the error from above using the fact that $x_i \in (0,\frac 12)$. You won't get this fast a decrease, but any factor less than $1$ is good enough.

Ross Millikan
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  • Ok. I'm happy that you understand my struggle. I wasn't sure if I expressed myself clear enough with my problem. But you confirm that there is no way to build an equality with $0$ in that place (it would be a different function, so I can't just analyse what is 'easier to analyse'), right? – theboyboy Nov 24 '20 at 06:13
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    I don't understand what you mean by build an equality with $0$ in some place. I am subtracting off the limit from a term, so you could look at this as a new series of the $\epsilon_i$, which will converge to $0$. There are other ways to show convergence than a bounded monotone sequence. Sequences that bounce above and below the limit but keep getting closer converge as well. An approach like mine will work with those. The multiplier on $\epsilon$ comes out negative, but greater than $-1$ so the distance from the limit is always reduced. – Ross Millikan Nov 24 '20 at 06:17
  • I was not referring directly to your answer, but to my question, to get a definitive answer. Just wanted to confirm that there is no way to make here $x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$ an equality with $0$ and analyse simpler function. If there is no way, then I will use your method for sure. – theboyboy Nov 24 '20 at 06:23
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This is a Möbius transformation. Once you get the roots $l_1, l_2$ of the characteristic function $l^2+2l-1=0$, it follows that $1-2l_1=l_1^2$ and $1-2l_2=l_2^2$. Then

$$ x_{n+1}-l_1 = \frac{1}{2+x_n}-l_1 = \frac{1-2l_1-l_1 x_n}{2+x_n} = \frac{l_1^2-l_1 x_n}{2+x_n} = -l_1 \frac{x_n-l_1}{2+x_n} \tag 1 $$

Similarly $$ x_{n+1}-l_2 = -l_2 \frac{x_n-l_2}{2+x_n} \tag 2 $$

$(1) \div (2)$ (you can do this because $x_n>0>l_2$), $$ \frac{x_{n+1}-l_1}{x_{n+1}-l_2} = \frac{l_1}{l_2}\cdot \frac{x_n-l_1}{x_n-l_2} $$

Therefore $\frac{x_n-l_1}{x_n-l_2}$ is a geometric sequence,

$$ \frac{x_n-l_1}{x_n-l_2} = \left(\frac{l_1}{l_2} \right)^{n-1} \cdot \frac{x_1-l_1}{x_1-l_2} \tag3 $$

Then $$x_n=\frac{l_1-\frac{x_1-l_1}{x_1-l_2}\left( \frac{l_1}{l_2}\right)^{n-1} \cdot l_2}{1- \frac{x_1-l_1}{x_1-l_2}\left(\frac{l_1}{l_2}\right)^{n-1}}$$

As $n\to \infty, \left(\frac{l_1}{l_2}\right)^{n-1} \to 0, x_n \to l_1 = \sqrt 2 - 1$.

To solve using matrices, see here for an example.

Neat Math
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