The Unapologetic Mathematician

Mathematics for the interested outsider

Convergence of Complex Series

Today, I want to note that all of our work on convergence of infinite series carries over — with slight modifications — to complex numbers.

First we have to get down an explicit condition on convergence of complex sequences. In any metric space we can say that the sequence z_n converges to a limit z if for every \epsilon>0 there is some N so that d(z_n,z)<\epsilon for all n>N. Of course, here we’ll be using our complex distance function |z_n-z|=\sqrt{(z_n-z)\overline{(z_n-z)}}. Now we just have to replace any reference to real absolute values with complex absolute values and we should be good.

Cauchy’s condition comes in to say that the series \sum\limits_{k=0}^\infty a_k converges if and only for every \epsilon>0 there is an N so that for all n\geq m>N the sum \left|\sum\limits_{k=m}^na_k\right|<\epsilon.

Similarly, we say that the series \sum\limits_{k=0}^\infty a_k is absolutely convergent if the series \sum\limits_{k=0}^\infty|a_k| is convergent, and this implies that the original series converges.

Since the complex norm is multiplicative, everything for the geometric series goes through again: \sum\limits_{k=0}^\infty c_0r^k=\frac{c_0}{1-r} if |r|<1, and it diverges if |r|>1. The case where |r|=1 is more complicated, but it can be shown to diverge as well.

The ratio and root tests are basically proven by comparing series of norms with geometric series. Since once we take the norm we’re dealing with real numbers, and since the norm is multiplicative, we find that the proofs go through again.

August 28, 2008 Posted by | Analysis, Calculus, Power Series | 1 Comment

   

Follow

Get every new post delivered to your Inbox.

Join 393 other followers