could someone please explaine to me how his power series has a radus of convergence and what it is:
the sum between n=0 and infinity of (n!)^2 Z^n. when Z member of the complex plain
Well, every power series has a radius of convergence!
To find the radius of convergence of $\displaystyle \sum_{n=0}^\infty (n!)^2 z^n$, use the "ratio test" (that or, occaisionally the root test, is almost always the best way to find radii of convergence).
$\displaystyle |\frac{a_{n+1}}{a_n}|= \left|\frac{((n+1)!)^2 z^{n+1}}{n!)^2 z^n}\right|= \frac{n+1}{n}\frac{n+1}{n} |z|$.
In order for the power series to converge that must go to a number less than 1 as n goes to infinity. In fact, it goes to infinity for all non-zero z but for z= 0, it is just 0.
That is, this series, like every power series, has a "radius of convergence" but it is 0! The series converges for z= 0 but no other z.