# Complex Power Series/Examples/3^n-1 over 2^n+1

## Example of Complex Power Series

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

- $a_n = \dfrac {3^n - 1} {2^n + 1}$

The complex power series:

- $S = \displaystyle \sum_{n \mathop \ge 0} a_n z^n$

has a radius of convergence of $1$.

## Proof

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

By Radius of Convergence from Limit of Sequence:

- $R = \displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }$

Thus:

\(\displaystyle a_n\) | \(=\) | \(\displaystyle \dfrac {3^n - 1} {2^n + 1}\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\frac {3^n} {2^n} - \frac 1 {2^n} } {1 + \frac 1 {2^n} }\) | $\quad$ multiplying top and bottom by $\dfrac 1 {2^n}$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {\frac 3 2 - \frac 1 {2^n} } {1 + \frac 1 {2^n} }\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(\to\) | \(\displaystyle \dfrac {3 / 2} 1\) | $\quad$ as $n \to \infty$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac 3 2\) | $\quad$ and so is independent of $n$ | $\quad$ |

Thus:

\(\displaystyle \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n - 1} } {a_n} }\) | \(=\) | \(\displaystyle \cmod {\dfrac {3 / 2} {3 / 2} }\) | $\quad$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) | $\quad$ | $\quad$ |

$\blacksquare$

## Sources

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 4$. Elementary Functions of a Complex Variable: Exercise $3 \ \text {(ii)}$