Continued fraction of binomial function $(1+z)^{1/4}$

Question

If I ask WolframAlpha for the continued fraction of $(1+z)^{1/4}$, I get

and I wonder what would be the representation for any $k = n$? Could you give me any hints or a reference?

Answer 1

This is a particular case of Gauss' continued fraction for the ratio of hypergeometric functions:

$$\frac{{_2 F_1}(A,B+1;C+1;Z)}{{_2 F_1}(A,B;C;Z)}=\cfrac{1}{1-\cfrac{\frac{A(C-B)}{C(C+1)}Z}{1-\cfrac{\frac{(B+1)(C-A+1)}{(C+1)(C+2)}Z}{1-\cfrac{\frac{(A+1)(C-B+1)}{(C+2)(C+3)}Z}{1-\cfrac{\frac{(B+2)(C-A+2)}{(C+3)(C+4)}Z}{1- \dots}}}}}$$

The choice of functions gets a little tricky, so I will just write without proof (see also the article for the hypergeometric function):

$$(1+z)^a=1+az \frac{{_2 F_1}(1,1+a;2;-z)}{{_2 F_1}(1,a;1;-z)}$$

Substituting the parameters, we obtain:

$$(1+z)^a=1+\cfrac{az}{1+\cfrac{\frac{1 \cdot (1-a)}{1 \cdot 2}z}{1+\cfrac{\frac{1 \cdot (1+a)}{2 \cdot 3}z}{1+\cfrac{\frac{2 \cdot (2-a)}{3 \cdot 4}z}{1+\cfrac{\frac{2 \cdot (2+a)}{4 \cdot 5}z}{1+ \dots}}}}}$$

I hope it's clear how to continue.

Now substitute $a=\frac{1}{4}$ to get your continued fraction.