# 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.

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You want to have $\limsup_{n \to \infty} \sqrt[n]{|a_n|} \lt 1$. See Wikipedia. Could you be more specific on what sort of decay property you are looking for? – t.b. Jun 10 '11 at 14:09
@Theo Buehler: does faster than $\frac{1}{n^p}$ for any $p \in \mathbb{N}$ sufficient ? – Rajesh Dachiraju Jun 10 '11 at 14:13
@Rajesh: but is then for p=1 the series not divergent ( harmonic series)? – Gottfried Helms Jun 10 '11 at 14:20
@Gottfried Helms : I mean faster than the case of any $p \in \mathbb{N}$, i do not mean $p = 1$, i mean for any arbitrarily large $p$. – Rajesh Dachiraju Jun 10 '11 at 14:23
Why the -1 vote on this question? – Eric Naslund Jun 10 '11 at 15:39

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 $n$.

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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,

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