How do we know/prove a complex power series converges? and how do we get the domain of convergence?

Question

I'm looking up definitions but I fail to visualize this. I understand how to do Taylor/Laurent expansions and how to find the radius of convergence but I don't know if a series converges or not in the first place. I cannot visualize it when I look at its function. I haven't done it for real numbers because I come from the French system where we haven't done Taylor expansions in high school contrary to the American system, I just learnt this now in my complex analysis class but I'm lost. Any general help to understand convergence would be appreciated!

Answer 1

If it helps with the understanding, I think an intuitively understanding is good.

We know by definition that the Taylor series is equal to the power series at $x_0=0$, however, if $x_0\neq 0$ one can just substitute. I give an example;

We have the function $f(x)=x^{-1}$, which can be written as: $$f(x)=\frac{1}{x}=\frac{1}{1-(1-x)}=\sum_{n=0}^\infty(1-x)^n$$ for $|1-x|<1$, as it is a geometric series then. Which we can write as (which is the power series): $$\sum_{n=0}^\infty (-1)^n(x-1)^n.$$ We can then define $y=x-1$, so $$f(y+1)=\sum_{n=0}^\infty (-1)^n y^n$$ Where the last then is the Taylor series to the power series we had preivous, as the last one is evaluated at $x_0=0=y+1$.

With the above understanding, we know that a complex power series can be expressed as: $$\sum_{n=0}^\infty a_n(z-z_0)^n$$ on the domain $B=\left\{z||z-x_0|<r\right\}$. We know that it can only converge if it has the radius $r$, as the sum will just diverge. This can be seen together with the relationship between Taylor and power series.

To find the domain, one can often use the quotient test, if it is positive series, Leibniz + some other test, if alternating, or just Cauchy-Hellman, to find the domain of convergence. There are many methods, but these are those I most often see.