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$$\operatorname{exp}(z)=\sum_{n=0}^\infty \frac{z^n}{n!}$$

This converges absolutely for every $z\in \Bbb C$. What does this mean to a layman?

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  • $\begingroup$ Can you define a layman? Is it someone with no mathematics background at all? $\endgroup$
    – Ducky
    Feb 14, 2015 at 12:59
  • $\begingroup$ What would a layman need complex numbers, exponentials, and power-series for? $\endgroup$ Feb 14, 2015 at 12:59
  • $\begingroup$ Hmmm. I have done heaps of math, but I don't know what converging absolutely means, especially on complex numbers $\endgroup$ Feb 14, 2015 at 13:00
  • $\begingroup$ The layman is me if you want to just have a quick look at my past questions $\endgroup$ Feb 14, 2015 at 13:00
  • $\begingroup$ Are you asking the definition of 'absolute convergence'? $\endgroup$
    – Git Gud
    Feb 14, 2015 at 13:02

1 Answer 1

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The ratio from one term to the next is $z/n$. Eventually, $n$ is bigger than $|z|$, so the terms from that point on get smaller. Eventually, $n>2|z|$, so the terms halve in size each step. So the sum will converge. That is true for any $z$.
We don't have the problem of $\frac1{1-z}=1+z+z^2+z^3+...$ which stops converging when $|z|\geq1$.

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  • $\begingroup$ Thank you, that makes sense $\endgroup$ Feb 14, 2015 at 13:10
  • $\begingroup$ You haven't addressed absolute convergence, though. $\endgroup$
    – Pedro
    Feb 14, 2015 at 13:22
  • $\begingroup$ I did when the terms halved in size each step. $\endgroup$
    – Empy2
    Feb 14, 2015 at 13:23

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