# A series expansion for \$left(x^2,B(x,x)right)^frac nx\$

Imagine we wanted to find a series expansion for $$left(x^2,B(x,x)right)^frac nx$$ which would be particularly difficult because it uses the Beta function. Let there be the rule that any strategies can be used to find a series expansion for $$0le xle 1$$ as long as the series terms are not too complicated and have some pattern. Here is how I would do it using a (Taylor)(expansion(https://en.m.wikipedia.org/wiki/Taylor_series) expansion at x=a which all give beautiful results. One simple and amazing expansion would be at x=0, but to include all cases, I will use x=a as all cases for $$0le ale 1$$ have the radius of convergence which works for $$0le xle 1$$:

$$left(x^2,B(x,x)right)^frac nx =frac{x!^frac{2n}{x}}{Γ(2x)^frac nx}=Γ^{-frac nx}(2x)sum_{n=0}^infty frac{left(frac{d^n}{dx^n}x!^frac {2n}xright)_{x=a}(x-a)^n}{n!}$$

Here comes the problem where I am unsure about the radius of convergence for the usual Taylor series of $$Γ(2x) ^{-frac nx}$$. The problem is this steep change due to the x in the denominator of $$left(frac{(2x)!}{2x}right)^{-frac nx}$$. If it was the factorial instead of the gamma, there would not be such a small derivative at $$x≈0$$ which would imply the series representation working. For example here is the graph of $$frac{1}{Γ(2x)^frac 1x}$$ for $$n=1$$:

This all relates to the series of $$x^frac 1x$$ of which I still do not know the convergence for. It amazingly has an explicit form for the Taylor Series at $$x=1$$ at OEIS A008405. How can I do the rest or have a new series?

It does not have to be a Taylor Series.

Please correct me and give me feedback!