I need to find a Taylor Series expansion of $\displaystyle \int_{0}^{z}e^{\zeta^{2}}d\zeta$ around $z=0$, which shouldn't be hard enough.
Except that I can only integrate term-by-term if the Taylor $\displaystyle e^{\zeta^{2}}$ is uniformly convergent. I don't think it is, because by the Weierstrass M-Test,each of the $\displaystyle e^{\zeta^{2}}$ needs to be bounded, by some $M_{n}$ with $n$ exceeding some fixed number $N>0$, and they're not.
Unless there's some other way to show that $\displaystyle e^{\zeta^{2}}$ is uniformly convergent? Or unless I'm not doing the Weierstrass M-test right?
Please let me know how to justify this integration! (And it must be justifiable, because our prof. wouldn't have given us this problem if it weren't!)
$$\mathrm e^{\zeta^2} = 1 + \zeta^2+\frac{1}{2!}\zeta^4+\frac{1}{3!}\zeta^6+\cdots+\frac{1}{n!}\zeta^{2n}+\cdots$$
If $|\zeta| \le 1$ then $$\left|\frac{1}{n!}\zeta^{2n}\right| \le \frac{1}{n!} = M_n$$ It is well known that $$1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}+\cdots = \mathrm e < \infty$$ This show that the series for $\mathrm e^{\zeta^2}$ converges uniformly on $|\zeta| \le 1$.