Complex Sine Function is Entire/Proof 1

Theorem

Let $\sin: \C \to \C$ be the complex sine function.

Then $\sin$ is entire.

Proof

By the definition of the complex sine function, $\sin$ admits a power series expansion about $0$:

$\ds \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$

By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\sin$ is entire it suffices to show that this series is everywhere convergent.

From Radius of Convergence from Limit of Sequence: Complex Case, it is sufficient to show that:

$\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 3}!} \times \frac {\paren {2 n + 1}!} {\paren {-1}^n} } = 0$

We have:

\(\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 3}!} \times \frac {\paren {2 n + 1}!} {\paren {-1}^n} }\)

\(=\)

\(\ds \size {-1} \lim_{n \mathop \to \infty} \size {\frac {\paren {2 n + 1}!} {\paren {2 n + 3} \paren {2 n + 2} \paren {2 n + 1}!} }\)

Definition of Factorial

\(\ds \)

\(=\)

\(\ds \lim_{n \mathop \to \infty} \paren {\frac 1 {\paren {2 n + 3} \paren {2 n + 2} } }\)

\(\ds \)

\(=\)

\(\ds 0\)

Hence the result.

$\blacksquare$