condition for the radius of convergence of a power series to be greater than 1 Consider the power series 
$F(z) = \sum\limits_{n=1}^{\infty} a_n z^n$
my question is how should the $a_n$ decay for the radius of convergence of the series to be greater than 1.
 A: Hint: 
What happens if $a_n=\frac{1}{2^n}$?  Consider the series $$\sum_{n=1}^\infty  \frac{z^n}{2^n}.$$  Does it converge if $z=1$?  What about $z=1.5$?  And $z=2$?
Also, here is another series:  Let $a_n=\frac{1}{n!}$, so we are looking at $\sum_{n=1}^\infty \frac{z^n}{n!}$.  It turns out that this series will converge for every $z\in\mathbb{C}$.  Try to use the ratio test to prove why.
Hope that helps,
A: As Theo said it:

The radius of convergence is greater than $1$ if and only if $\limsup\sqrt[n]{|a_n|}<1$. 

In turn, this condition is met if and only if there exists a finite $c$ and a real number $r<1$ such that $|a_n|\le cr^n$ for every $n$. 
The condition that $|a_n|\le cn^{-p}$ is not sufficient, for any positive $p$, it only ensures that the radius of convergence is at least $1$ but not that it is greater than $1$. Even the condition that $n^p|a_n|\to0$ for every given $p$ is not enough. Consider for example the sequence $a_n=\exp(-(\log n)^2)$, then $\sqrt[n]{|a_n|}\to1$ because $(\log n)^2\ll n$ hence the radius of convergence of the associated series is $1$ although $n^p|a_n|\to0$ for every given $p$ because $p\log n\ll(\log n)^2$. 
In fact, powers of $n$ are on a too fine grained scale for radiuses of convergence: to wit, for every sequence $(a_n)$, the series $\displaystyle\sum a_nz^n$ and $\displaystyle\sum a_nn^pz^n$ have the same radius of convergence, for every real number $p$.
Likewise, for every finite positive $R$, the radius of convergence is greater than $R$ if and only if there exists a finite $c$ and a real number $r<1/R$ such that $|a_n|\le cr^n$ for every $n$.
