3
$\begingroup$

Let $a_1=1$ and $a_{n+1}=1+\frac{1}{a_n}.$

Is $a_n$ convergent?

How could i find its limit?

I found even terms of the sequence decrease and odd terms are increase. But i cant find upper and lower bounds to use monotone convergence theorem.

$\endgroup$
  • 6
    $\begingroup$ Here is one hint: If the limit exists (call it $A$), then $A=1+1/A$. Can you see why? Oh, and one more question: Was a value for $a_1$ (or $a_0$) given? $\endgroup$ – Harald Hanche-Olsen Dec 6 '14 at 11:50
  • 2
    $\begingroup$ $a_1=1$ Ok but how could i find the sequences $(a_{2n})$ and $(a_{2n+1})$ approaches to same limit? $\endgroup$ – user35 Dec 6 '14 at 11:52
  • 6
    $\begingroup$ You could show the sequence is contractive: $|a_{n+2}-a_{n+1}|\le C|a_{n+1}-a_n|$ with $0<C<1$. (Note this, by itself, would imply the sequence converges.) $\endgroup$ – David Mitra Dec 6 '14 at 11:58
  • 2
    $\begingroup$ As for the bounds: there is an obvious lower bound. Once you've found that, there is an almost as obvious upper bound. $\endgroup$ – David Mitra Dec 6 '14 at 12:04
  • $\begingroup$ first prove $a_n\in(1,2)$ then we can use high-low limit to prove it's convergent. $\endgroup$ – Idele Dec 7 '14 at 8:27
3
$\begingroup$

Who knows derived following criteria may apply:

If $f:I \rightarrow I$ is differentiable and $p$ is a fixed point such that $f'(p) <1$ then the sequence given by $x_{n+1}=f(x_n)$ este converges to $p$.

For $$f:(0,\infty) \rightarrow (0,\infty),f(x)=1+\frac{1}{x},f'(x)=-\frac{1}{x^2},p=\frac{1+\sqrt{5}}{2}, |f'(p)|=\frac{2}{3+\sqrt{5}}<1.$$

$\endgroup$
2
$\begingroup$

You can prove by induction that $$a_n = \frac{F_{n+1}}{F_n}$$ where $\{F_n\}_{n\geq 1}=\{1,1,2,3,5,8,\ldots\}$ is the Fibonacci sequence.

Binet formula hence gives: $$\lim_{n\to +\infty} a_n = \phi = \frac{1+\sqrt{5}}{2}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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