Weierstrass M Test and Analytic Functions I have studied how to solve some PDEs using Fourier's Method. But, I'd like to know why don't we just use Taylor Series to solve these. After a little research, I found two interesting results: 1st) Uniform convergence Vs. Pointwise Convergence, which I understood fairly fine; 2nd) Cauchy-Kowaleski Theorem, which I could not.
My attempt to solve, generally, DEs (via Taylor Series) seems to be wrong according to the 1st topic. That's because I can only ensure pointwise convergence. Here's what I have been doing so far: I substitute every function into its Taylor Series form. Then, I conclude it will converge if:
$$\lim_{n \to \infty} \left| \frac{f^{(n+1)}(z_0)}{f^{(n)}(z_0)} \right| \leq M \in \mathbb{R}_+\tag{1}\label{1}  $$
In order to achieve \eqref{1}, I rewrite $f$ into its Taylor Series form and then apply the ratio test. However, as it seems, it will ensure pointwise convergence which it's not sufficient.
So, my question is: May we use Weierstrass M-Test to check whether or not the Taylor Series will converge uniformly? If so, how should I proceed with the computations? Let's assume all functions $f$ are complex analytic in their open and simply connected domains.
How am I able to find every $M_n$ (https://en.wikipedia.org/wiki/Weierstrass_M-test) ?
Any help and recommendations are welcome. Thanks
 A: Power series don't converge uniformly on their disc of convergence. But they do so locally, meaning that they converge uniformly on every compact subset of their disc of convergence. The Weierstraß-M-test can be used to prove this.
First notice that every compact subset of a disc centered at $z_0$ is contained in a compact disc also centered at $z_0$. So it's sufficient to show uniform convergence on compact discs $K_r(z_0)$ with radius $r$ smaller than the radius of convergence $R$. On such a disc we have
$$\vert a_k(z-z_0)^k\vert=\vert a_k\vert\vert z-z_0\vert^k\leq\vert a_k\vert r^k,$$
and the series with these terms is known to converge because the original power series converges absolutely. Now we can apply the test to show uniform convergence on the compact disc. But it won't generally converge on the entire disc of convergence. For instance, the series expansion of the exponential function doesn't. Neither does the series expansion of $\frac{1}{1-x}$ at $0$.
A: What I think you're missing is the definition of the radius of convergence.

*

*A power series $\sum_{n=0}^\infty a_n z^n $ converges absolutely and pointwise for $|z|<R$, where $R$ is the radius of convergence, defined by
$$ \frac1R \overset{\triangle}= \limsup_{n\to\infty} |a_n|^{1/n}.$$
Indeed, for each $z$ such that $|z|<R$, we have $|z|/R<1$, so we can use the geometric series formula as follows:
$$ \sum_{n=0}^\infty |a_n z^n| = \sum_{n=0}^\infty (|a_n|^{1/n} |z|)^n \le  \sum_{n=0}^\infty \left (\frac1R|z|\right)^n = \frac1{1-|z|/R} < \infty. $$


*Once you know that it converges on some open ball $|z|<R$, you can check that  for each $\epsilon>0$, it converges uniformly on the set $\{z:|z|<R-\epsilon\}$. This is by the Weierstrass M test:
$$ |a_n z^n| \le \left(\frac1R|z|\right)^n \le \left(\frac{R-\epsilon}R\right)^n =: M_n $$
since $\sum_{n=0}^\infty M_n < \infty$ (again, geometric series), $\sum_{n=0}^\infty a_n z^n$ converges absolutely and uniformly.
