Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
1  
You do know that you don't necessarily need the solution in order to check for convergence, right? –  user1337 Jul 3 '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 3 '13 at 18:35
    
And that means $x_{n+2}$ is larger than ... –  Daniel Fischer Jul 3 '13 at 18:35
    
Check the convergence of monotone sequences. –  Mhenni Benghorbal Jul 3 '13 at 18:37
    
@MhenniBenghorbal This sequence isn't monotone. If you consider every other term, however, then it is. –  Arthur Jul 3 '13 at 18:38

5 Answers 5

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.

share|improve this answer

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 $\lim_{k \to \infty} x_{n_k}$.
share|improve this answer

Try

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

share|improve this answer
    
These look like Fibonacci numbers to me! –  Ali Jul 3 '13 at 19:07

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.

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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