<s>[INST] Find MacLaurin Series and radius of convergence $$\sin{2z^2}$$
$$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$
Let's replace z with $2z^2$ in the MacLaurin series:
$$\sin{2z^2} = \sum_{n=0}^{\infty}(-1)^n \frac{(2z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n 
 \frac{2^{2n+1}}{(2n+1)!} z^{4n+2}$$
$$= z^2 - \frac{2^3}{3!}z^6 + \frac{2^5}{5!}z^{10} -+ ... = z^2 - \frac{4}{3}z^6 + \frac{4}{15}z^{10} -+...$$
But what about the radius of convergence? How do I find this?
 [/INST] Isn’t it quite easy? Start with the standard Laurent series
$$
\sin(z)=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\,,
$$
make your substitution $z\to2z^2$, and get
$$
\sin(2z^2)=\sum_{n=0}^\infty(-1)^n\frac{2^{2n+1}z^{4n+2}}{(2n+1)!}\,,
$$
in which your $a_n$-term is $(-1)^n2^{2n+1}z^{4n+2}/(2n+1)!$.
Now form $\mid a_{n+1}/a_n\mid$ to get
$$
\left|\frac {a_{n+1}}{a_n}\right|=\frac{2^{2n+3}z^{4n+6}\big/(2n+3)!}{2^{2n+1}z^{4n+2}\big/(2n+1)!}
=\frac{(2n+1)!}{(2n+3)!}\frac{2^{2n+3}z^{4n+6}}{2^{2n+1}z^{4n+2}}=\frac{4z^4}{(2n+2)(2n+3)}\,.
$$
That rightmost thing clearly goes to zero, no matter what $z$ is, so your radius is infinite, as I’m sure you expected.
</s>