Complex Sine Function is Entire/Proof 1

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:

hence the result.