Check: Radius of Convergence of the Sum of these Complex Taylor Series I just found the following Taylor series expansions around $z=0$ for the following functions:


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*$\displaystyle \frac{1}{z^{2}-5z+6} = \frac{1}{(z-2)(z-3)} = \frac{-1}{(z-2)} + \frac{1}{(z-3)} = \sum_{n=0}^{\infty}\frac{z^{n}}{2^{n+1}} - \sum_{n=0}^{\infty}\frac{z^{n}}{3^{n+1}} = \sum_{n=0}^{\infty}\left( \frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}\right)z^{n}$

*$\displaystyle \frac{1}{1-z-z^{2}} = \frac{1}{\left[z-\left(\frac{1-\sqrt{5}}{-2} \right) \right]\left[z-\left(\frac{1+\sqrt{5}}{-2} \right)\right]} = \frac{2\sqrt{5}}{5(1-\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1-\sqrt{5}} \right)z} - \frac{2\sqrt{5}}{5(1+\sqrt{5})}\frac{1}{1-\left(\frac{-2}{1+\sqrt{5}} \right)z} = \frac{\sqrt{5}}{5} \sum_{n=0}^{\infty}\left(\frac{(-1)^{n}2^{n+1}}{(1-\sqrt{5})^{n+1}} - \frac{(-1)^{n}2^{n+1}}{(1+\sqrt{5})^{n+1}} \right)z^{n}$
I wanted to confirm that the radius of convergence for the first series is $|z|<2$ and that the radius of convergence for the second series is $|z|<\frac{-1+\sqrt{5}}{2}$.
I know that for a series $\sum_{n=1}^{\infty}a_{n}z^{n}$ with radius of convergence $R_{1}$ and $\sum_{n=1}^{\infty}b_{n}z^{n}$ with radius of convergence $R_{2}$, the radius of convergence of the series $\sum_{n=1}^{\infty}(a_{n}+b_{n})z^{n}$ is some $R$ where $R \geq \min(R_{1}, R_{2})$. And it can even be the case that if both of the series have finite $R_{1}$ and $R_{2}$, that $R$ can be infinite! 
So, I wanted to make sure this was not the case here, and that I have the correct radii of convergence for both. If not, how do I go about finding them (preferably in the least icky way possible)?
Thank you.

You can use Hadamard's rule:
$$\frac 1R=\limsup_n a_n^{1/n}=\limsup_n \biggl(\frac1{2^{n+1}}-\frac1{3^{n+1}}\biggr)^{\!1/n}\!.$$
Now rewrite this expression as
$$\frac1{2^{(n+1)/n}}\biggl(1-\frac{2^{n+1}}{3^{n+1}}\biggr)^{\!1/n}=\frac1{2^{1+1/n}}\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{1/n}\!.$$
The first factor tends to $\frac12$. As to the second factor,
$$\frac23<1-\Bigl(\frac23\Bigr)^{n+1}<1,\enspace\text{whence}\quad\Bigl(\frac23\Bigr)^{1/n}<\biggl(1-\Bigl(\frac23\Bigr)^{n+1}\biggr)^{\!1/n}<1, $$
and by the squeeze principle, it tends to $1$. Thus $\limsup_n a_n^{1/n}=\frac12$, and $R=2$.
