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How can we find the value to which the following series converges, if it converges to a finite number? If else, how can we prove that it is divergent?$$1+\frac{1}{2+\frac{1}{3+......}}$$

Did
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vamsi3
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  • Can you see it as a sequence defined by induction? – paf Aug 20 '16 at 18:04
  • No. I can't. I don't get your idea. – vamsi3 Aug 20 '16 at 18:06
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    http://mathworld.wolfram.com/ContinuedFractionConstant.html Apparently, this is a number simply called the "continued fraction constant." – florence Aug 20 '16 at 18:08
  • @user104014: it was a wrong idea. – paf Aug 20 '16 at 18:12
  • @florence Oh! Thanks for the link. It completely explains all my doubts. – vamsi3 Aug 20 '16 at 18:22
  • If it converges, it is towards an irrational number, cause the development of a rational number is periodic. – Sylvain Julien Aug 20 '16 at 18:54
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    @SylvainJulien All simple continued fractions converge. The ones which are (eventually) periodic are roots of a quadratic, the ones which are finite, ie zero after a certain point are rational. – snulty Aug 20 '16 at 19:02
  • Thank you snulty for the information. – Sylvain Julien Aug 20 '16 at 19:06
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    @SylvainJulien It's actually covered quite well in hardy and Wright. Or I'm sure it's in a lot of number theory books. The proof essentially shows that out of the odd and even terms one increases and the other one decreases (I can't remember which) and both have to meet at some limit. – snulty Aug 20 '16 at 19:09
  • @florence this one seems to be one over the continued fraction constant – snulty Aug 20 '16 at 19:31
  • @MJD Why adding the huge vertical space in a title? – Did Aug 20 '16 at 20:42

1 Answers1

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We can see that $$ 1 < 1+\frac{1}{2+\frac{1}{3+......}} $$ and $$ \frac{1}{2} > \frac{1}{2+\frac{1}{3+......}} $$ so $$ 1 < 1+\frac{1}{2+\frac{1}{3+......}} < \frac{3}{2} $$

HiPocketRockets
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  • you show that it is bounded but that does not imply that it converges – miracle173 Aug 20 '16 at 19:09
  • @miracle173 It would be easier or just as easy to show all simple continued fractions converge $\ldots$ – snulty Aug 20 '16 at 19:11
  • It's bounded and we can define a sequence which decreases and get's asymptotically closer to it: $s_1=\frac{1}{1+0}$, $s_2= \frac{1}{2+\frac{1}{1}}$ and so on... – Squirtle Aug 20 '16 at 19:29
  • @Squirtle it doesn't decrease, it alternates above and below its limit. Check the next few terms – snulty Aug 20 '16 at 19:50