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This is just a curiousity question.

HISTORY:-

$Motivation:$

I first started wondering about this question about 3 months ago when I first got interested in continued fractions but I really didn't tried to get its value but when I saw the thumbnail of this video https://youtu.be/CaasbfdJdJg, I quickly clicked and after watching the whole video I realized that the thumbnail was a click bait and the host didn't even talked about the thing in the thumbnail.

$Introduction:$

After the biggest anime betrayal of my life from one of my favourite math channel, I tried to get the value of $1+\frac{1}{2+\frac{1}{3+\dots}}$ by myself even though I didn't really have enough mathematical knowledge and as you would expect I really got nowhere and after some sad math noises I forgot about the expression for nearly 2 months but suddenly yesterday I remembered the thing again and here I am asking this question.

THE NUMBER:-

Let $\textstyle\displaystyle{Q=1+\frac{1}{2+\frac{1}{3+\ddots}}}$

Obviously whatever $Q$ is, it is an irrational number because the continued fraction is infinite.

After playing with wolfram alpha it is clear that $1.4<Q<1.5$ is true.

It also seems compelling that $Q$ does converge to a specific real value because

If we define $\textstyle\displaystyle{Q_n=1+\frac{1}{2+\frac{1}{3+\dots\frac{1}{n}}}}$

Then, $\textstyle\displaystyle{Q=\lim_{n\rightarrow\infty}(Q_n)}$

$\textstyle\displaystyle{|Q_1-Q_2|=\frac{1}{2}}$

$\textstyle\displaystyle{|Q_2-Q_3|=\frac{1}{14}}$

$\textstyle\displaystyle{|Q_3-Q_4|=\frac{1}{210}}$

So it seems, $\textstyle\displaystyle{\lim_{n\rightarrow\infty}(Q_n-Q_{n-1})=0}$

Finally the question:-

I am not much of a mathematician, so please for the mathematician folks out there I would like to know the closed form of $Q$ if there exists one and also the derivation of it or some bounding. (The previous bounding was not derived but rather observed from wolfram alpha)

Star Alpha
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Rounak Sarkar
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    Hope you learnt the lesson and read articles in the future, wikipedia for a start and to get deeper PDF-files. If I remember right, this number has something to do with the bessel-function. Surely, someone will soon give you the suitable reference. – Peter Feb 19 '21 at 13:17
  • Why don't you give the references? – Rounak Sarkar Feb 19 '21 at 13:18
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    The decimal expansion of your number is http://oeis.org/A060997 (its reciprocal is http://oeis.org/A052119). Its relation to Bessel functions is explained in https://mathworld.wolfram.com/ContinuedFractionConstants.html. – player3236 Feb 19 '21 at 13:19
  • @RanjitKumarSarkar I do not remember enough details to quickly find the reference, but you got it now anyway. – Peter Feb 19 '21 at 13:20
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    Yep, it's the ratio of modified Bessel functions of the first kind. Pretty impossible to guess, but the proof comes from recurrence relations which exist for Bessel functions. A pretty interesting topic – Yuriy S Feb 19 '21 at 13:23
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    Also see https://math.stackexchange.com/questions/1871733/convergence-of-a-harmonic-continued-fraction. – player3236 Feb 19 '21 at 13:23
  • Thanks to everyone for helping. – Rounak Sarkar Feb 19 '21 at 13:28
  • Despite all my years playing with Bessel functions, I had no idea this relation existed. Even experts on something learn something new every day! – Cameron Williams Feb 19 '21 at 13:30
  • Glad that you learned something new from my question but did you laughed while reading my question? – Rounak Sarkar Feb 19 '21 at 13:39
  • The answer is actually given in a reply to one of the top comments: https://www.youtube.com/watch?v=CaasbfdJdJg&lc=Ugz6-ld_pMpXdiujTpB4AaABAg.94YLzb3CL0e9EKDXSAhYEt – Hans Lundmark Feb 19 '21 at 16:11
  • By the way, all simple continued fractions with positive integer terms converge, and quite rapidly. – MJD Oct 04 '21 at 14:29
  • @MJD. Where can I get the proof of that? – Rounak Sarkar Oct 04 '21 at 14:31
  • This is the Seidel-Stern theorem. You can find the proof in the book Continued Fractions by A. Ya. Khinchin (University of Chicago press, 1964). It is theorem 10 on page 10. – MJD Oct 04 '21 at 15:11
  • The proof is just a generalization of the argument you used for this special case. – MJD Oct 04 '21 at 15:26

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