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I have the complex functions $f_n(z) = 1/(1+z^n)$ and I'm supposed determine where $\sum_{n=1}^\infty f_n(z)$ converges for $z \in\mathbb{C}$

Extra Info:
I was only able to determine convergence for $|z| \gt 1$. The argument applies the ratio test:
$|\frac{f_{n+1}}{f_n}| = \frac{|1 + z^n|}{|1+z^{n+1}|}$
Now note that the denominator $|1 + z^{n+1}| = |1 -(-z^{n+1})| \ge |1 - |-z^{n+1}|| = |1 - |z|^{n+1}| \ge \frac{1}{K}|z|^{n+1}$ for large $n$ and $K \gt 1$
So then the ratio can be bounded
$\frac{|1 + z^n|}{|1+z^{n+1}|} \le K\frac{1 + |z|^n}{|z|^{n+1}} = \frac{K}{|z|^{n+1}} + \frac{K}{|z|}$ which limits to $\frac{K}{|z|}$ so convergence is guaranteed by ratio test when $|z| \gt K$ and $K$ was arbitrarily greater than $1$

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1 Answer 1

up vote 2 down vote accepted

For any point $z \in \bar{\mathbb D}$ you can apply the divergence test to your summation to conclude that it does not converge.

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Oh duh! For some reason I thought that the divergence test didn't apply here because z is complex. –  Mark Feb 19 '13 at 6:09

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