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Would someone be able to verify that my proof that the following sequence converges is sound?

We have a sequence $\left ( x_{n} \right )_{n\geq i}$ where $x_{1} = 3$ and $x_{n+1}= \frac{1}{4 - x_{n}}$. We wish to show that this sequence converges (and then determine its limit). So, I applied the monotone convergence theorem, showing it is bounded and monotone decreasing by induction and hence converges.

So, the base case is fine, showing that $\frac{1}{4} \leq x_{2} = 1 < x_{1} = 3 \leq 3$. Then I made the assumption that $\frac{1}{4} \leq x_{k+1} < x_{k} \leq 3$. Now, we want to show:

$\frac{1}{4} \leq x_{k+2} < x_{k+1} \leq 3$ (I think this is correct?)

So, it is probably possible to do this as one big inequality if you are good at that, however, for simplicity, I split it and showed $x_{k+2} < x_{k+1}$ first and then that $\frac{1}{4} \leq x_{k+2}$ and $x_{k+1} \leq 3$.

So:

$x_{k+1} < x_{k}$ (by our assumption)

$ \Rightarrow 4 - x_{k} < 4 - x_{k+1} $

$ \Rightarrow \frac{1}{4 - x_{k+1}} < \frac{1}{4 - x_{k}} $ (both terms less than 3 by our assumption, so, we can divide without changing inequality sign)

$ \Rightarrow x_{k+2} < x_{k+1} $

$ \Rightarrow$ monotone decreasing

So, now that we have shown that the sequence is decreasing, we just need to show both bounds, like so:

$\frac{1}{4} \leq x_{k+1}$ and $x_{k} \leq 3$ (by our assumption)

$ \Rightarrow 4 - x_{k+1} \leq 4 - \frac{1}{4}$ and $ 4 - 3 \leq 4 - x_{k}$

$ \Rightarrow \frac{1}{\frac{15}{4}} \leq \frac{1}{4-x_{k+1}} $ and $ \frac{1}{4 - x_{k}} \leq 1$

$ \Rightarrow \frac{1}{4} < \frac{4}{15} \leq x_{k+2}$ and $ x_{k+1} \leq 1 < 3$

Thus, combining both parts of the proof together once again we form the inequality:

$ \frac{1}{4} < x_{k+2} < x_{k+1} < 3$

Hence, by induction, we have shown that the sequence is bounded below by $\frac{1}{4}$, is strictly decreasing and is bounded above by 3, thus, the sequence converges by the monotone convergence theorem. Does anyone spot any errors in my proof?

Thanks!

Helen Byrne
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  • Looks fine in general but there are some minor points we must pay attention to: (1) in your proof $;x_{k+2}<x_{k+1};$ , when you do the step $$4-x_k<4-x_{k+1}\implies\frac1{4-x_{k+1}}<\frac1{4-x_k}$$ you're implicitly assuming these quantities are positive. You should perhaps at least mention this, or better: prove it. (2) If you've already show the sequence's monotone decreasing, all you need for the limit to exist is to show it is bounded from below, so don't waste time showing it is bounded from above, as it clearly is by $;x_1;$ ! – Timbuc Jan 07 '15 at 18:00
  • But isn't this the point of an induction proof? That we ASSUME that $x_{k}$ and $x_{k+1}$ are bounded by $\frac{1}{4}$ and 3, and so then in our inductive proof we can assume that this is true, so 4 - $x_{k}$ has to be between 1 and $\frac{15}{4}$ and hence we can divide knowing that these are in fact positive, because this is our inductive argument? – Helen Byrne Jan 07 '15 at 19:44
  • Indeed so, and that's why I wrote "to mention this" The "to prove it" comes from the first case, certainly. Very good post, worked out nicely. +1 – Timbuc Jan 07 '15 at 20:19
  • Thank you! Yes, I would certainly make sure to explain everything when writing it as an answer, to ensure I would get all of the marks, I was just keeping it a bit more concise here :) Thanks again! – Helen Byrne Jan 07 '15 at 20:32
  • Does this answer your question? Is this recursively defined sequence decreasing? $x_{n+1}={1\over{4-x_n}}$, $x_1=3$. – Anne Bauval Oct 22 '23 at 05:00

1 Answers1

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The simplest criterion to know if a sequence defined by a relation $x_{n+1}=f(x_n)$ is monotonic is to determine if all terms of the sequence belong to an interval on which $f(x)>x$ or $<x$.

Anyway, the limits, if any, are fixed points of the function $f$. Solving for $fx)=x$, one finds $ l_l, l_2 = 1\pm\dfrac{\sqrt{3}}2 <4$. As $f$ is increasing, one checks easily that

  • $f(]l_1,l_2[)\subset\,\, ]l_1,l_2[$,
  • $f(x)<x$ on $]l_1,l_2[$.

Hence the sequence is decreasing (trivial induction) and bounded below by $l_1$, so that it converges to $\,\,l_1=1-\dfrac{\sqrt{3}}2$.

Bernard
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  • I think the lecturer wants us to use the monotone convergence theorem, he hasn't shown us that method of showing it is monotonic and part (a) of the question was to state the monotone convergence theorem. However, thanks for explaining this alternative method! – Helen Byrne Jan 07 '15 at 19:46
  • @Helen Byrne: this ‘method of showing the sequence is monotonic’ is just a matter of common sense: if the sequence is defined by $x_{n+1}=f(x_n)$, saying the sequence is decreasing is equivalent to saying $f(x)<x$, and it it can been graphically: the representative vurve of $f$ i under the first bissectrix. – Bernard Jan 07 '15 at 22:51