I have to calculate the following limit using power series, but I'm really bad at it and don't even know how to start. Could someone please explain in detail? also what's the difference between a power series and a Taylor expansion?
$$\lim_{z \to 0} \frac{e^{z^2}-1}{(e^z-1)^2}$$
First we observe that we can expand $e^z$ in a power series: $e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$ which implies that $e^z-1=\sum_{n=1}^\infty \frac{z^n}{n!}$ and $e^{z^2}-1=\sum_{n=1}^\infty \frac{z^{2n}}{n!}.$ Now we can compute the limit
$$\lim_{z\to 0} \frac{e^{z^2}-1}{(e^z-1)^2}=\lim_{z\to 0} \frac{\sum_{n=1}^\infty \frac{z^{2n}}{n!}}{(\sum_{n=1}^\infty \frac{z^n}{n!})^2}=\lim_{z\to 0}\frac{z^2(\sum_{n=0}^\infty \frac{z^{2n}}{n!})}{z^2(\sum_{n=0}^\infty \frac{z^n}{n!})^2}=\lim_{z\to 0}\frac{1+(\sum_{n=1}^\infty \frac{z^{2n}}{n!})}{1+(\sum_{n=1}^\infty \frac{z^n}{n!})^2}=1$$
Might be easier to see if you don't look with the "sum" notation and open the sum instead.