Yes, there is. Consider a continued fraction in the form:

$$x=\cfrac{a}{b+\cfrac{a}{b+\cfrac{a}{b+\cdots}}}$$

Assume the limit exists and find it:

$$x=\cfrac{a}{b+x}$$

$$x^2+bx-a=0$$

$$x=\frac{\sqrt{b^2+4a}-b}{2}$$

Now consider the nested radical:

$$x=\sqrt{c+d\sqrt{c+d\sqrt{c+\cdots}}}$$

Assume the limit exists and find it:

$$x=\sqrt{c+dx}$$

$$x^2-dx-c=0$$

If we set $d=-b$ and $c=a$ we get exactly the same value of the limit. I assumed that $b>0$ so in this case $d<0$ and we get the radical:

$$x=\sqrt{a-b\sqrt{a-b\sqrt{a-\cdots}}}=\cfrac{a}{b+\cfrac{a}{b+\cfrac{a}{b+\cdots}}}$$

With the condition $a>b>0$, of course.

This is the most simple connection we could find, but of course there may be countless others.