Confusion on complex power series

Question

Let $f(z)=\sum_{j=0}^{\infty}{a_jz^j}$ be a power series centered at $0$ with radius of convergence $\alpha>0$, where $a_j,z \in\mathbb{C}$. In addition, denote the $n$th partial sum as $P_n(z)$.

I know that if $|a_n/a_{n+1}| \to \alpha$, then the radius of convergence of $\sum_{j=0}^{\infty}{a_jz^j}$ is $\alpha$. (http://en.wikipedia.org/wiki/Radius_of_convergence#Finding_the_radius_of_convergence)

But I can't seem to answer the following:

1. Given a power series $\sum_{j=0}^{\infty}{a_jz^j}$ with a radius of convergence of $\alpha$, does it follow that $|a_n/a_{n+1}|\to \alpha$?
2. If $|a_n/a_{n+1}| \to \alpha$, does it follows that ${P_n(z)/(a_nz^n)}\to z/(z-1)$ uniformly (or pointwise) for $|z|>\alpha$?

For 1, I can't seem to convince myself that this is true. And for 2, I've made little progress. I'm having difficulty controlling the quantity $|P_n(z)/(a_nz^n)- z/(z-1)|$. For sequence of real functions I can inspect $\sup |f_n-f|$...but in this case I'm hitting a brick wall. All I could come up with was \begin{align*} \bigg|\frac{P_n(z)}{a_nz^n} - \frac{z}{z-1}\bigg| &\le \bigg|\frac{P_n(z)}{a_nz^n}\bigg| + \bigg|\frac{z}{z-1}\bigg| \\ &\le \sum_{j=0}^{\infty}{\bigg|\frac{a_{n-j}}{a_n} \bigg||(1/z)|^j} + \bigg|\frac{z}{z-1}\bigg| \\ &\le (\alpha+\epsilon)\bigg|\frac{z}{z-1}\bigg| + \bigg|\frac{z}{z-1}\bigg| \end{align*}

Any ideas?

Answer 1

For 1. consider $\sum_{k=0}^\infty z^{2k}$. (Note that $a_n = 0$ if $n$ is odd.)

For 2. see what happens in a simple case, for example $\sum_{k=0}^\infty 2^k z^k$.