TITLE: Convergence of a Harmonic Continued Fraction
QUESTION [14 upvotes]: 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.

REPLY [2 votes]: 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.