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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.

Michael Hardy
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Confuse
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4 Answers4

31

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.

Michael Hardy
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Jack D'Aurizio
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8

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:

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

Yuriy S
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    How did you know that for the continued fraction to diverge, $ a_1 +a_2 + a_3 + \cdots $ must diverge? – GohP.iHan Jul 27 '16 at 11:16
  • @GohP.iHan, Seidel-Stern theorem, see for example http://math.stackexchange.com/q/556114/269624 P.S. You meant to say for the continued fraction to converge, right? – Yuriy S Jul 27 '16 at 11:21
  • @You'erInMyEye: Oh right. I miswrote that. Thanks for the link, but unfortunately, the link in that top solution is now dead. – GohP.iHan Jul 28 '16 at 08:37
  • 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 – Dabed Dec 08 '20 at 23:50
  • @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" – Yuriy S Dec 09 '20 at 12:48
  • I did before exactly the term you suggest and a whole lot of variations to no avail but will keep trying, if you remember it please let me know, thank you very much. – Dabed Dec 09 '20 at 15:07
  • @DanielD., of course, don't mention it. If it helps, I checked this numerically in Mathematica (due to slow convergence I had to use a lot of terms, but the value seems correct) – Yuriy S Dec 09 '20 at 16:09
  • that is nice, as it appears on wikipedia I just mindlessly assumed it was true but I should have verified it too, besides google I used approach0.xyz and found two mentions on MSE but no proof, I think I will have to post a question but as this contains the reciprocal of the naturals and because of the harmonics series and all that it looks weird to me that a proof of this doesn't pop up everywhere. – Dabed Dec 09 '20 at 17:14
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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 $1$.

Robert Israel
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2

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.

Sidharth Ghoshal
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