$$\operatorname{exp}(z)=\sum_{n=0}^\infty \frac{z^n}{n!}$$
This converges absolutely for every $z\in \Bbb C$. What does this mean to a layman?
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asked Feb 14, 2015 at 12:55
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$\begingroup$ Can you define a layman? Is it someone with no mathematics background at all? $\endgroup$– DuckyFeb 14, 2015 at 12:59
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$\begingroup$ What would a layman need complex numbers, exponentials, and power-series for? $\endgroup$– Hagen von EitzenFeb 14, 2015 at 12:59
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$\begingroup$ Hmmm. I have done heaps of math, but I don't know what converging absolutely means, especially on complex numbers $\endgroup$– Display NameFeb 14, 2015 at 13:00
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$\begingroup$ The layman is me if you want to just have a quick look at my past questions $\endgroup$– Display NameFeb 14, 2015 at 13:00
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$\begingroup$ Are you asking the definition of 'absolute convergence'? $\endgroup$– Git GudFeb 14, 2015 at 13:02
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1 Answer
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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 $|z|\geq1$.
