How can I show that tha radius of convergence for complex sine is $R = \infty$? In my notes it is stated that for $f(z) = \sin(z): \mathbb{C} \to \mathbb{C}$, defined as $\sin(z) =  \sum_{n = 0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!} $, the radius of convergence is obviously $R = \infty$.
However, I don't really see why this is true. The only theorem regarding radius of convergence that we've stated is the Hadamard's formula:
$$ \frac{1}{R} = \limsup_{n \to \infty} \sqrt[n] { |a_n| }$$
This formula isn't very useful here, because it requires me to evaluate a limit of a factorial under the $n$-th root, which is certainly not an easy job. If only we had the quotient test! Which we do, but only when the limit exists; if the limit belowe exists, then we have
$$ \frac{1}{R} = \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|} $$
Unfortunately, the limit does not exist, because all even coefficients are zero, while odd coefficients are of the form $ \frac{(-1)^k}{(2k+1)!} $. So, what am I missing to see the obvious truth that $R = \infty$?
 A: We can go about it a lot easier:
$$
\sum_{n = 0}^\infty\left \lvert (-1)^n \frac{z^{2n+1}}{(2n+1)!} \right \rvert = \sum_{n = 0}^\infty\left \lvert \frac{z^{2n+1}}{(2n+1)!} \right \rvert \leq \sum_{n = 0}^\infty \left \lvert \frac{z^n}{n!} \right \rvert
$$
You had alread realized that we are only dealing with odd powers. We have just added the even ones. Then:
$$
\sum_{n =0}^\infty \left \lvert \frac{z^n}{n!} \right \rvert = \sum_{n = 0}^\infty \frac{\lvert z \rvert^n}{n!} = \exp(\lvert z \rvert)
$$
Even if you do not know that this is $\exp$, it is very easy to prove that the last power series converges.
And if you really want to brute force the Hadamard formula, look up Stirling's formula which tells you how the factorial behaves.
A: It is doable with Hamadard's formula: notice that
$$(2n+1)! > n^n$$
so that
$$\sqrt[n] {|a_n|} = \sqrt[n] \frac{1}{(2n+1)!} < \sqrt[n] {n^{-n}} = 1/n \to 0.$$
The ratio test also works, if you consider the fractions $$\frac{|a_{n+2}|}{|a_n|}$$
instead. It is only an unimportant detail that you're running into, one can prove that the ratio test also works in this situation.
A: You've cited a consequence of the ratio test that is usually helpful for power series but isn't helpful for $\sin z$. However, the original ratio test itself works perfectly! (Steven alluded to his in their answer) In the series
$$
\sin z = \sum_{k=0}^\infty \frac{(-1)^k z^{2k+1}}{(2k+1)!},
$$
the ratio of the modulus of successive terms is
$$
\frac{\bigl|(-1)^{k+1} z^{2(k+1)+1}/(2(k+1)+1)!\bigr|}{\bigl|(-1)^k z^{2k+1}/(2k+1)!\bigr|} = \frac{|z^2|}{(2k+2)(2k+3)};
$$
since this tends to $0$ as $k\to\infty$ (no matter what $z$ is) and $0<1$, the ratio test implies that the series converges (no matter what $z$ is).
