# Complex Power Series/Examples/2n Factorial over n Factorial Squared

## Example of Complex Power Series

Let $\sequence {a_n}$ be the sequence defined as:

$a_n = \dfrac {\paren {2 n}!} {\paren {n!}^2} z^n$
$S = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$

has a radius of convergence of $\dfrac 1 4$.

## Proof

Let $R$ denote the radius of convergence of $S$.

Thus:

 $\displaystyle R$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }$ Radius of Convergence from Limit of Sequence $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {\paren {2 \paren {n - 1} }!} {\paren {\paren {n - 1}!}^2} / \dfrac {\paren {2 n}!} {\paren {n!}^2} }$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {\paren {2 n - 2}!} {\paren {2 n}!} \dfrac {\paren {n!}^2} {\paren {\paren {n - 1}!}^2} }$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {n^2} {\paren {2 n} \paren {2 n - 1} } }$ simplifying $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {n^2} {4 n^2 - 2 n} }$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac 1 {4 - \frac 2 n} }$ $\displaystyle$ $=$ $\displaystyle \frac 1 4$ as $\sequence {\dfrac 1 n}$ is a basic null sequence

$\blacksquare$