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$$x_{n+1} = \frac {1}{x_n + 1}; x_1 > 0$$

How to transform it into the form $x_n = $? I need the solution in order to check if it converges at any $x_1 > 0$.

John
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    You do know that you don't necessarily need the solution in order to check for convergence, right? – user1337 Jul 03 '13 at 18:33
  • No matter what you put in there for $x_n$, it's pretty obvious that $x_{n+1}$ is less than $1$. At least once it's been pointed out. As for convergence, a limit $x$ needs to statisfy $$ x = \frac{1}{x + 1} $$ since it can't be a limit unless it's a stationary point. Now just solve that equation and check that for any $x_n$, $x_{n+1}$ is closer to this $x$. – Arthur Jul 03 '13 at 18:35
  • And that means $x_{n+2}$ is larger than ... – Daniel Fischer Jul 03 '13 at 18:35
  • Check the convergence of monotone sequences. – Mhenni Benghorbal Jul 03 '13 at 18:37
  • @MhenniBenghorbal This sequence isn't monotone. If you consider every other term, however, then it is. – Arthur Jul 03 '13 at 18:38
  • This sequence can be represented as a continued fraction as $n\to\infty$ by $[0;1,1,1,1,1,…]$ which is essentially $\phi^{-1}$. – Maazul Jul 03 '13 at 18:40
  • @Panda, no. Then what should I do? – John Jul 03 '13 at 18:41
  • @Arthur, the roots are $-0.5 \pm \frac {\sqrt 5} {2}$. How to know which on is limit? – John Jul 03 '13 at 18:43
  • @John Try a few (very) different values for $x_0$, calculate up to $x_4$ or something, see if you see any patterns. It's how any mathematician would do it if he didn't see a solution right away. – Arthur Jul 03 '13 at 18:43
  • @John Can any term of the sequence ever be negative? Positive? How does that help you choose? – Arthur Jul 03 '13 at 18:43
  • @Arthur, if the 1st term is positive, then all subsequent are positive. As a result. there is no need to check :) – John Jul 03 '13 at 18:51
  • @John Mind you, you've just figured out what the limit has to be if it exists. But knowing your limits is half the battle. – Arthur Jul 03 '13 at 18:55
  • Also, if you are just interested in the closed form solution check here: http://www.wolframalpha.com/input/?i=x%5Bn%2B1%5D%3D1%2F%281%2Bx%5Bn%5D%29 – Ali Jul 03 '13 at 19:10

5 Answers5

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hint

One nice way to do linear fractional recurrences is to use matrices. If $a/b=x$, then $c/d = 1/(x+1)$, where $$ \left(\begin{aligned}0\qquad 1 \cr 1\qquad 1\end{aligned} \right) \left(\begin{aligned}a\cr b\end{aligned}\right) = \left(\begin{aligned}c\cr d\end{aligned}\right) $$ So we can get a formula for $x_n$ if we know a formula for the $n$th power of that $2 \times 2$ matrix.

GEdgar
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Some hints:

  1. First of all you can prove by induction that all of the $x_n$'s are strictly positive.
  2. Using that, prove that they are also lesser than 1.
  3. From the above two steps, the sequence $\{ x_n \}$ lies in the compact interval $[0,1]$,so it has a convergent subsequence $\{ x_{n_k} \}$
  4. Prove that $\lim_{n \to \infty} x_n$ exists and is equal to $\lim_{k \to \infty} x_{n_k}$.
user1337
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Try

$$a(n)\to \frac{\mathcal C \;\left(\frac{1}{2} \left(1+\sqrt{5}\right)\right)^n+\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^n}{\mathcal C \; \left(\frac{1}{2} \left(1+\sqrt{5}\right)\right)^{n+1} +\left(\frac{1}{2} \left(1-\sqrt{5}\right)\right)^{n+1}}$$

al-Hwarizmi
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Let $a=\lim_{n\to\infty}x_n$. Then

\begin{align} &a=\frac{1}{1+\frac{1}{1+\frac{1}{1+...}}}\\&a=\frac{1}{1+a}\\&a(a+1)=1\\&a^2+a-1=0\\&a=\frac{-1\pm\sqrt{5}}{2} \end{align}

Since there is no negative term or subtraction in the sequence, we omit the negative solution.

$\displaystyle \therefore a=\lim_{n\to\infty}x_n=\frac{\sqrt{5}-1}{2}$

Maazul
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If you are interested in finding the limit, assume $\lim_{n\to \infty} a_n = a$, then

$$ a=\frac{1}{a+1} \implies a^2+a-1=0 \implies a=\frac{-1 \bar{+} \sqrt{5}} {2}. $$

Now, you should pick up the right limit.

Mhenni Benghorbal
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