Find radius of convergence of the power series $\frac{\text{sin}^2(z)}{z}$

Question

$\textit{Problem:}$

Find radius of convergence of the power series $\frac{\text{sin}^2(z)}{z}$.

$\textit{Solution}$: We need to write $ \frac{\text{sin}^2 (z)}{z}$ as a series of powers, and then calculate their convergence radius. In effect, note that:

$\text{sin}^2(z)=\frac{1-\text{cos}(2z)}{2}=2^{-1}-2^{-1}\sum_{n=0}^\infty\frac{(-4)^nz^{2n}}{(2n)!}\\=2^{-1}-\sum_{n=0}^\infty\frac{(-1)^n2^{2n-1}z^{2n}}{(2n)!}=-\sum_{n=1}^\infty\frac{(-1)^n2^{2n-1}z^{2n}}{(2n)!}$

Then: $\frac{\text{sin}^2(z)}{z}=\sum_{n=1}^\infty\frac{(-1)^{n+1}2^{2n-1}z^{2n-1}}{(2n)!}=\sum_{n=0}^\infty\frac{(-1)^{n+2}2^{2n+1}z^{2n+1}}{(2(n+1))!} \\=\sum_{n=0}^\infty\frac{2(-4)^{n}z^{2n+1}}{(2(n+1))!}$

How can I calculate the convergence radius of this series? Thanks for your attention.

Answer 1

Let $F\colon\mathbb C\longrightarrow\mathbb C$ be the function defined by$$F(z)=\begin{cases}\frac{\sin^2z}z&\text{ if }z\neq0\\0&\text{ if }z=0.\end{cases}$$Then $F$ is analytic and your power series is the Taylor series of $F$ centered at $0$. But then its radius of convergence is $\infty$, because the Taylor series of an analytic function centered at a point $z_0$ always converges on any disk centered at $z_0$.

Answer 2

HINT

Use the fact that $$\sqrt[n]{n!} \to +\infty$$

Note that $(2(n+1))! \geq n!$