2
$\begingroup$

I would like some advice on how to solve problems like the following: Let $(x_n)$ be a sequence defined by $x_1= 3$ and $x_{n+1} = \frac{1}{4-x_n}$. Prove that the sequence converges.

My strategy is to use the Monotone Convergence Theorem, but I am having trouble showing that the sequence is decreasing and bounded below. Here's my work so far:

Decreasing: The first 4 values are $3,1,1/3,3/11$, so let's assume $x_n \leq x_{n-1} \leq \ldots \leq x_1$. Want to show $x_{n+1} \leq x_n$. We have $x_{n+1} = \frac{1}{4-x_n}$. I want to have an upper bound for the RHS, but can't find one and don't really know where to go from here.

Bounded below: I wanted to show that all values are positive, but if we assume $x_n > 0$, that doesn't rule out $x_{n+1}$ from being negative.

$\endgroup$
  • $\begingroup$ To prove the values are positive, it is easier to prove that all values are in the interval $(2-\sqrt{3}, 2+\sqrt{3})$. $\endgroup$ – vadim123 Apr 8 '16 at 23:48
  • $\begingroup$ It's not apparent to me why anyone would think to use that approach without already knowing something about this sequence. $\endgroup$ – rorty Apr 8 '16 at 23:54
  • $\begingroup$ You are lead to that interval by looking at fixed points. That is, if $f(x)=\frac 1{4-x}$ then $f(x)=x\implies x=2\pm \sqrt 3$. So...if the series converges, the limit is one of those two values. $\endgroup$ – lulu Apr 8 '16 at 23:58
1
$\begingroup$

assume that it convergences to x. In which case, $x = \frac{1}{4-x}$ solve for x. There are two possible solutions... one of those will prove to be unstable.

Now that you know what x should equal it should be simple to show that $x_1>x_2>...>x$ your sequence is bounded below and monotonically decreasing and therefore convergent.

Suppose {$x_n$} converges to $x$.

$x = \frac{1}{4-x}\\ 4x - x^2 = 1\\ - (x^2 -4x + 1) =0\\ - (x - 2+\sqrt3)(x - 2-\sqrt3)=0\\ $

I claim that $\forall x_n\in(2-\sqrt3,2+\sqrt3), 2-\sqrt3<x_{n+1}<x_n$

to show that $x_{n+1}<x_n$, we can show that $x_n - x_{n+1}>0$

$x_n - x_{n+1} = x_n - \frac{1}{4-x_n} = \frac{-x_n^2 + 4x_n - 1}{4-x_n} $ and $\frac{-x_n^2 + 4x_n - 1}{4-x_n} > 0$ when x is in the interval.

And to show that $x_{n+1} > 2-\sqrt3$, when $x_n>2-\sqrt3, 4-x_n < 2+\sqrt3$ and $\frac{1}{4-x_n} > \frac{1}{2+\sqrt3}.$

$x_2 \in(2-\sqrt3,2+\sqrt3)$ then for all $n>2, 2-\sqrt3<x_{n+1}<x_n$

$x_n$ is monotonically decreasing and is bounded below.

$\endgroup$
  • $\begingroup$ Allowing $x = \frac{1}{4-x}$ is a heuristic just to lead to possible solutions right? I can't see how you can rigorously allow that equation from the definition of the limit. Does this usually work with recursively defined sequences? $\endgroup$ – rorty Apr 9 '16 at 0:12
  • $\begingroup$ If the sequence has a limit $L$, then $L$ must satisfy $L=\frac{1}{4-L}$. $\endgroup$ – vadim123 Apr 9 '16 at 0:16
  • $\begingroup$ Indeed. We make a leap of faith that this is the lower bound, and then we can rigorously show that it in fact is. $\endgroup$ – Doug M Apr 9 '16 at 0:16
  • $\begingroup$ @vadim123 I now see your point. What you say is true because $\lim x_n = \lim x_{n+1}$ combined with the algebraic limit theorem. $\endgroup$ – rorty Apr 9 '16 at 0:58
  • $\begingroup$ @DougM Can you elaborate on how knowing the limit helps make the argument that the sequence is decreasing? I found that the sequence is bounded below. $\endgroup$ – rorty Apr 9 '16 at 1:14

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.