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I am working on a multi-part problem, which shows that $$\sqrt{2} = 1+\frac{1}{2+\frac{1}{2+...}}$$

I have shown that given some rational approximation $\frac{m}{n}$ to $\sqrt2$ , we can take $\frac{m'}{n'} = \frac{m+2n}{m+n}$ and get a better approximation.

I also know that this iteration goes back and forth between being larger and smaller than $\sqrt2$. Say if we start with $\frac{m}{n}<\sqrt2$, then iterating will give us $$\frac{m}{n}<\frac{m''}{n''} = \frac{3m+4n}{2m+3n}<\sqrt2$$

Finally, I am asked to prove that the sequence obtained by iterating in this fashion, starting with $$1,\frac{3}{2},\frac{7}{5},...$$ which I have shown is given recursively by $$q_1=1, q_{n+1} = 1+\frac{1}{1+q_n} $$

converges to $\sqrt2$.

My professor's "hint" is to consider the "odd" and "even" subsequences, and show that they both converge to the same limit. This makes sense to me, since both sequences are clearly monotone and bounded. Most (all?) of my classmates simply wrote that because they are both monotone and bounded by $\sqrt2$, then they must both converge to $\sqrt2$. However, just because they are bounded above and below, respectively by $\sqrt2$, it does not necessarily mean that $\sqrt2$ is the limit for both (or either!) of them. Some classmates have tried to prove by contradiction that if there were some other limit for one of the subsequences, we could iterate again and get closer to $\sqrt2$. I think that this is a faulty argument, because it assumes that we reach the limit. Remember: the sequence is "fixed" as soon as we choose to start it at $q_1 = 1$ !

My counterargument towards most of my classmates has been the sequence given my $\frac{1}{n}$. The sequence is bounded below by $-99$, and no matter how many times we "iterate" (in this case, just taking the next value of $n$) we can always get closer and closer to $-99$, but that doesn't mean that $-99$ is the limit.

If someone could offer me a resolution to this problem I would be hugely appreciative. It's been driving me crazy for the past week.

EDIT: I realize this post has been marked as a duplicate, however I think it should be left up, as the answer given is a bit different from that in the "original".

NNN
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  • just because they are bounded above and below ... does not necessarily mean that sqrt(2) is the limit for both You are right, that needs to be proved. Try to show that the differences between consecutive terms tend to $0$ for example. – dxiv Sep 30 '16 at 01:51
  • @WillJagy I am not exactly sure what you mean by "convergents for $\sqrt2$". – NNN Sep 30 '16 at 02:02
  • LaTeX note: use \cfrac for continued fractions instead of ordinary \frac. It will typeset nicer. – Oiler Sep 30 '16 at 02:03
  • @dxiv We have not yet covered Cauchy sequences, so I don't think that is an acceptable answer unfortunately. – NNN Sep 30 '16 at 02:04
  • @Alex My hint was essentially equivalent to David's answer. You don't need Cauchy for that. Once you prove that the difference between consecutive terms vanishes to $0$, it follows that if both even and odd subsequences converge, then the difference between their limits is $0$. – dxiv Sep 30 '16 at 02:07

2 Answers2

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Note that $$\Bigl|\frac{m'}{n'}-\frac mn\Bigr| =\Bigl|\frac{m+2n}{m+n}-\frac mn\Bigr| =\Bigl|\frac{m^2-2n^2}{(m+n)n}\Bigr|\ .$$ In the same way $$\Bigl|\frac{m''}{n''}-\frac{m'}{n'}\Bigr| =\Bigl|\frac{(m')^2-2(n')^2}{(m'+n')n'}\Bigr|\ ;$$ substituting back in terms of $m,n$ and doing a bit more algebra gives $$\Bigl|\frac{m''}{n''}-\frac{m'}{n'}\Bigr| =\Bigl|\frac{m^2-2n^2}{(2m+3n)(m+n)}\Bigr| <\frac13\Bigl|\frac{m'}{n'}-\frac mn\Bigr|\ .$$ Thus the difference between successive approximations approaches zero; since, as you have already proved, odd and even terms are alternately less than and greater than $\sqrt2$, each must approach $\sqrt2$.

David
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  • This is a wonderful answer! Thanks so much! How did you come up with this? I couldn't imagine ever having thought of this. – NNN Sep 30 '16 at 02:10
  • Can't really claim credit, sadly... it's a well-known technique in continued fractions. If you would like to read further there are many sources, including my own lecture notes at http://www.maths.unsw.edu.au/~angell/5535/chapter4.ps – David Sep 30 '16 at 02:29
  • Sorry to bother again, but could you explain how we know that

    $$\lvert{\frac{m^2-2n^2}{(2m+3n)(m+n)}\rvert < \frac{1}{3}\lvert\frac{m'}{n'} - \frac{m}{n}\rvert$$

    Sorry. I cannot seem to get the formatting correct in this comment. I'm try to ask how you know that the last inequality involving 1/3 is true.

     – NNN Sep 30 '16 at 03:07
  • $2m+3n>3n$, then compare the first line. – David Sep 30 '16 at 03:12
  • Wonderful. Thanks so much for your patience! – NNN Sep 30 '16 at 03:28
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$$ \begin{array}{cccccccccccccccccccccccccccccc} & & 1 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & & 2 & \\ \frac{0}{1} & \frac{1}{0} & & \frac{1}{1} & & \frac{3}{2} & & \frac{7}{5} & & \frac{17}{12} & & \frac{41}{29} & & \frac{99}{70} & & \frac{239}{169} & & \frac{577}{408} & & \frac{1393}{985} & & \frac{3363}{2378} & & \frac{8119}{5741} \\ \\ & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & & -1 \end{array} $$ Given the convergent $\frac{p}{q},$ what would you think is the $\pm 1$ directly underneath?

Will Jagy
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