Production expresion for Taylor series expansión.

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i need help, I am doing the Taylor series development around the origin of:

$\sqrt{z+i}$, i have considerated $\sqrt{z+i}=\sqrt{i} \sqrt{1+\frac{z}{i}}$ and make the variable change $x= \frac{z}{i}$ and doing the development in Taylor series of $\sqrt{1+x}$

But i have found that the expression of the nth derivate is:

$(\frac{1}{2}-0) (\frac{1}{2}-1) (\frac{1}{2}-2)... (\frac{1}{2}-(n-1))$

You know how i can get an expression for that since? i need to calculate the radius convergence.

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Just use the binomial theorem $$\sqrt{z+i}=\sum_{n=0}^\infty i^{\frac{1}{2}-n} \binom{\frac{1}{2}}{n} z^n$$