I know how to calculate the exact value for continued fractions such as
$$1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}=\frac{1+\sqrt{5}}{2}$$
However, is it possible to find the value of continued fraction
$$1+\cfrac{1}{2+\cfrac{1}{6+\cfrac{1}{24+\cfrac{1}{\ddots}}}}$$
($[1!;2!,3!,4!,5!,6!,\dots]$)
Thanks!
Best Answer
It converges, but not to anything particular at all.
Take a look at this