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

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!

• See "Finding the Radius of Convergence": en.wikipedia.org/wiki/Radius_of_convergence . Jun 11, 2018 at 2:03
• Thank you, I'm more confused about the domain of convergence though. Does it simply mean the disk of center z0 and radius the radius of convergence? Jun 11, 2018 at 2:08
• It depends on the point you pick. $$f(z)= \sum_{n \geq 0} a_n(z-b)^n$$ is a series about the point $b$, so if the radius of convergence is $R$, then $f$ converges for all points less than $R$ away from $b$. Jun 11, 2018 at 2:11
• You think American students know Taylor expansions from high school?
– zhw.
Jun 11, 2018 at 3:55

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.