Series expansion of $\sqrt{\log(1+x)}$ at $x=0$ Mathematica gives the following series expansion of $\sqrt{\log(1+x)}$ at $x=0$.
$$
x^{1/2}-\frac{1}{4}x^{3/2}+\frac{13}{96}x^{5/2}-\cdots
$$
You can find it from Wolfram alpha too.
How can I obtain the expansion?
Obviously Taylor expansion is impossible because $\sqrt{\log(1+x)}$ is not analytic at $x=0$.
Taylor expansion of the $\log(1+x)$ at $x=1$ is possible. But I don't know how to take sequre root on the expanded series.
I think I have not learned about square root of a series from calculs or analysis course.
From what material can I study about such things?

We have
$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
and recall that
$$(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)x^2}{2}+\frac{\alpha(\alpha-1)(\alpha-2)x^3}{3}+\cdots$$
so for $\alpha=\frac{1}{2}$ we have
$$\sqrt{\log(1+x)}=\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4)\right)^{1/2}=\sqrt{x}\left(1+(\underbrace{-\frac{x}{2}+\frac{x^2}{3}+O(x^3))}_{=u}\right)^{1/2}\\=\sqrt{x}(1+\frac{1}{2}u-\frac{1}{8}u^2+O(u^3))=\sqrt{x}(1-\frac{x}{4}+\frac{x^2}{6}-\frac{1}{8}\frac{x^2}{4}+O(x^3))\\=\sqrt{x}(1-\frac{x}{4}+\frac{13x^2}{96}+O(x^3))$$
