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Does this continued fraction converge?

$$\cfrac { 1 }{ 1+\cfrac { 1 }{ 2+\cfrac { 1 }{ 3+\cfrac { 1 }{ 4+\ddots } } } } $$

($[0;1,2,3,4, \dots]$)

I tried approximating a few values but I couldn't make out whether it converges or diverges. Can anyone provide a proof whether this converges or diverges. If it converges, please add what it converges to.

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    $\begingroup$ I like this question ;) It seems slightly related to one i asked, what have you tried so far? Anything other that approximating? $\endgroup$ Commented Jul 26, 2016 at 15:14
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    $\begingroup$ All simple continued fractions converge. $\endgroup$ Commented Jul 26, 2016 at 15:14
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    $\begingroup$ mathworld.wolfram.com/ContinuedFractionConstant.html $\endgroup$
    – Rob
    Commented Jul 26, 2016 at 15:18
  • $\begingroup$ ^ Might as well put that as the answer. $\endgroup$
    – Rellek
    Commented Jul 26, 2016 at 15:23
  • $\begingroup$ I was trying to find upper and lower bounds of the continued fraction. But the link makes it clear. Thanks. $\endgroup$
    – Confuse
    Commented Jul 26, 2016 at 15:24

4 Answers 4

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The modified Bessel functions of the first kind fulfills the recurrence relation: $$ I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{1}$$ due to the integral representation: $$ I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx \tag{2}$$ and the cosine addition formulas. A straightforward consequence of $(1)$ is that: $$ \frac{I_n(\alpha)}{I_{n-1}(\alpha)}=\cfrac{1}{\cfrac{2}{\alpha}\,n+\cfrac{1}{\cfrac{2}{\alpha}\,(n+1)+\cdots}}\tag{3} $$ hence by plugging in $\alpha=2$ and $n=1$ we get: $$\boxed{ \cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cdots}}}=\frac{I_1(2)}{I_0(2)}=\color{red}{\frac{\sum_{m\geq 0}^{\phantom{A}}\frac{1}{m!(m+1)!}}{\sum_{m\geq 0}\frac{1}{m!^2}}}\approx 0.697774657964.}\tag{4} $$ The convergence of the LHS is granted by the fact that it is an ordinary continued fraction, $[0;1,2,3,4,\ldots]$. Bounds are easily derived from the red ratio, since both the numerator and the denominator are very fast-converging series.

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    $\begingroup$ That is a nice one Jack (+1) $\endgroup$
    – tired
    Commented Jul 26, 2016 at 18:43
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    $\begingroup$ @tired: thank you, tired :) $\endgroup$ Commented Jul 26, 2016 at 18:44
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    $\begingroup$ It took me a while to figure out how to go from the recurrence to the continued fraction. This is pretty cool +1 $\endgroup$ Commented Dec 11, 2021 at 20:16
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By Seidel-Stern theorem, for a continued fraction of the following form to converge:

$$\cfrac { 1 }{ a_1+\cfrac { 1 }{ a_2+\cfrac { 1 }{ a_3+\cfrac { 1 }{ a_4+\dots } } } }$$

$$a_k >0$$

The following sum has to diverge:

$$a_1+a_2+a_3+a_4+\cdots$$

Thus, even the following continued fraction converges:

$$\cfrac { 1 }{ 1+\cfrac { 1 }{1/2+\cfrac { 1 }{ 1/3+\cfrac { 1 }{1/4+\dots } } } }=\frac{\pi}{2}-1$$

But this one diverges:

$$\cfrac { 1 }{ 1+\cfrac { 1 }{1/2^2+\cfrac { 1 }{ 1/3^2+\cfrac { 1 }{1/4^2+\dots } } } }$$

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    $\begingroup$ How did you know that for the continued fraction to diverge, $ a_1 +a_2 + a_3 + \cdots $ must diverge? $\endgroup$
    – GohP.iHan
    Commented Jul 27, 2016 at 11:16
  • $\begingroup$ @GohP.iHan, Seidel-Stern theorem, see for example math.stackexchange.com/q/556114/269624 P.S. You meant to say for the continued fraction to converge, right? $\endgroup$
    – Yuriy S
    Commented Jul 27, 2016 at 11:21
  • $\begingroup$ @You'erInMyEye: Oh right. I miswrote that. Thanks for the link, but unfortunately, the link in that top solution is now dead. $\endgroup$
    – GohP.iHan
    Commented Jul 28, 2016 at 8:37
  • $\begingroup$ How do you proof the continued fraction for pi/2-1? I have been looking for a proof but I can't find it and I don't see it myself either $\endgroup$
    – Dabed
    Commented Dec 8, 2020 at 23:50
  • $\begingroup$ @DanielD., I don't remember the details, but I believe we could transform it formally into a series and then prove that the series converges to this value. You could probably google "continued fractions for pi" $\endgroup$
    – Yuriy S
    Commented Dec 9, 2020 at 12:48
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This particular one is $I_1(2)/I_0(2)$ where $I_0$ and $I_1$ are modified Bessel functions of the first kind, of orders $0$ and $1$.

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    $\begingroup$ Hm... this answer feels... empty. $\endgroup$ Commented May 28, 2017 at 21:52
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Jack's answer is perfect but just in case you're as dumb as I am, here are the steps connecting the recurrence to the actual continued fraction:

$$ I_n(\alpha) = \frac{\alpha}{2n} \left( I_{n-1}(\alpha) - I_{n+1}(\alpha) \right) $$

$$ \frac{I_n(\alpha)}{I_{n-1}(\alpha)} = \frac{\alpha}{2n} \left( 1 - \frac{I_{n+1}(\alpha)}{I_{n-1}(\alpha)} \right) $$ $$ \frac{I_n(\alpha)}{I_{n-1}(\alpha)} = \frac{\alpha}{2n} \left( 1 - \frac{I_{n}(\alpha)}{I_{n}(\alpha)} \frac{I_{n+1}(\alpha)}{I_{n-1}(\alpha)} \right) $$

$$ \frac{I_n(\alpha)}{I_{n-1}(\alpha)} = \frac{\alpha}{2n} \left( 1 - \frac{I_{n}(\alpha)}{I_{n-1}(\alpha)} \frac{I_{n+1}(\alpha)}{I_{n}(\alpha)} \right) $$

Now we solve for $\frac{I_{n}}{I_{n-1}}$

$$ \left(1 + \frac{\alpha}{2n} \frac{I_{n+1}(\alpha)}{I_n(\alpha)} \right) \frac{I_n(\alpha)}{I_{n-1}(\alpha)} = \frac{\alpha}{2n} $$

$$ \frac{I_n(\alpha)}{I_{n-1}(\alpha)} = \frac{\frac{\alpha}{2n}}{\left(1 + \frac{\alpha}{2n} \frac{I_{n+1}(\alpha)}{I_n(\alpha)} \right)} $$

$$ \frac{I_n(\alpha)}{I_{n-1}(\alpha)} = \frac{\frac{2n}{\alpha}}{\frac{2n}{\alpha}} \frac{\frac{\alpha}{2n}}{\left(1 + \frac{\alpha}{2n} \frac{I_{n+1}(\alpha)}{I_n(\alpha)} \right)} = \frac{1}{\frac{2n }{\alpha} + \frac{I_{n+1}(\alpha)}{I_n(\alpha)}} $$

Now you can repeatedly substitute to yield the infinite continued fraction.

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    $\begingroup$ The $\frac{2\alpha}{n}$ at the very end should be $\frac{2n}{\alpha}$. $\endgroup$
    – Gary
    Commented Dec 11, 2021 at 23:34
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    $\begingroup$ thanks for pointing will fix when I get to a laptop. It is hard correct the LaTeX with my phone. $\endgroup$ Commented Dec 14, 2021 at 7:45

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