I have been asked to find the expansion (Taylor series) of $e^z$ at $z=0$. I do it and find that the answer is $$1+z+\frac{z^2}{2!}+\ldots$$ I find the radius of convergence using de alembert's ratio test: $$\lim_{n\to\infty}a_n/a_{n+1}.$$ Since $a_n=f^{(n)}(0)/n!$, I can easily calculate $a_n$ and $a_{n+1}$, since all derivatives of $e^z$ is $e^z$.
But what if a series like $\sin z$ is given? How to find $a_n$ and $a_{n+1}$? It depends on the value of $n$, which is nor known to us right? Or can we just take any $a_n$? Like if we do $\lim_{n\to \infty} a_2/a_3$?
When you repeatedly differentiate $\sin z$, you return to $\sin z$ after four steps. Use this recursively to evaluate $f^{(n)}(0)$.
When you try the ratio test on this: it fails, because some terms are $0$ and you cannot divide by them. So use the ratio test on the non-zero terms to determine the radius of convergence. Or use the root test, which does not require convergence, because you only need to use the $\limsup$.