Showing the following function is entire... The full problem asks about the following function using it's Maclaurin series:
$$f(x)=\left\{
\begin{array}{lr}
\frac{\sin(z)}{z} & : z \neq 0\\
\;\;\;\;1 & : z=0
\end{array}
\right.$$
I've represented $\dfrac{\sin(z)}{z}$ as the Maclaurin series
$$\frac{\sin(z)}{z}=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n+1)!}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-...$$
Applying the ratio test, we find that
$$\left|\lim_{n\to\infty}\frac{z^{2n+1}\cdot(2n+1)!}{z^{2n}\cdot(2n+3)!}\right|$$
$$=\left|\lim_{n\to\infty}\frac{z}{(2n+3)(2n+2)}\right|$$
$$=|z|\left|\lim_{n\to\infty}\frac{1}{(2n+3)(2n+2)}\right|$$
$$=|z|\cdot 0$$
Using the notation for radius of convergence as $R=\frac{1}{\beta}$, we have that $\beta=0$, so $R=\infty$. Does that mean that our function is entire since it converges everywhere? I mean, we know of a series that its derivative and integral will share the same radius of convergence, so then they in turn must be convergent everywhere. Now, I know when not using series, we want to satisfy the Cauchy-Riemann equations and look at if the function is analytic or not, so I figure looking at the derivative of our series and it being convergent everywhere entails entire. Anyone able to verify my thinking?
 A: On $\mathbb{C}\setminus \{0\}$, we claim that $f(z)=\frac{\sin z}{z}$ is analytic, you can use Morera's theorem to verify it, i.e for any simple close curve in $\gamma \in\mathbb{C}\setminus \{0\}$, we have $\int_\gamma \frac{\sin z}{z}=0$(No pole in the interior of $\gamma$), so $f(z)=\frac{\sin z}{z}$ is analytic on  $\mathbb{C}\setminus \{0\}$.
On $B(0,1)$, for $\lim_{z\rightarrow 0} \frac{\sin z}{z}=1$, $\frac{\sin z}{z}$ is bounded on $B(0,1)$, then we get $z=0$ is a removable singularity of $\frac{\sin z}{z}$, define its value at $z=0$ by $1$, i.e $$f(x)=\left\{
\begin{array}{lr}
\frac{\sin(z)}{z} & : z \neq 0\\
\;\;\;\;1 & : z=0
\end{array}
\right.$$
So we get $f$ is analytic on whole $\mathbb{C}$, namely an entire funciton.
By the way, as you have get the Laurent expansion of $f$ at $z=0$,
$$\frac{\sin(z)}{z}=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n}}{(2n+1)!}=1-\frac{z^2}{3!}+\frac{z^4}{5!}-..., \forall z\in \mathbb{C}\setminus \{0\}$$you can find its Laurent expansion is same as Talyor expansion(there is no term $\alpha_{-1}, \alpha_{-2},\cdots$), so we can say $f$ is analytic on whole $\mathbb{C}$.
