Convergence of a Harmonic Continued Fraction 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.
 A: 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:
$$\cfrac { 1 }{ 1+\cfrac { 1 }{1/2^2+\cfrac { 1 }{ 1/3^2+\cfrac { 1 }{1/4^2+\dots  }  }  }  }$$
A: 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.
A: 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 $1$. 
A: 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.
