Radius of Convergence of Power Series in Complex Plane

Theorem

Consider the complex power series:

$S = \ds \sum_{k \mathop = 0}^\infty z^n$

The radius of convergence $S$ is $1$.

Proof

\(\ds \lim_{n \mathop \to \infty} \dfrac {\cmod {z^n} } {\cmod {z^{n - 1} } }\)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {z^n} {z^{n - 1} } }\)

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \cmod z\)

\(\ds \)

\(=\)

\(\ds \cmod z\)

By the Ratio Test, it follows that:

$S$ is convergent for $\cmod z < 1$

$S$ is divergent for $\cmod z > 1$.

Hence the result by definition of radius of convergence.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4.4$. Power Series: Example $\text {(i)}$