The question is:

Prove or disprove:

The function $f(x)=x^{3/2}$ has a power series expansion about $x\_0=0$ with positive radius of convergence.

When I start calculating the derivative of the k'th order:

$f'(x)=3/2\cdot x^{1/2}$

$f''(x)=3/2\cdot 1/2 \cdot x^{-1/2}$

..

For every $K>=0$ derivative, $f^{k}(0)=0$

Does this mean, that this function doesn't have a taylor expansion? (and linking to this question - it doesn't have a radius of convergence) ?