# uniform convergence in complex plane

• Apr 21st 2010, 08:29 AM
mylestone
uniform convergence in complex plane
To show that $\sum_n^{\infty} \frac{z^n}{1-z^n}$ converges uniformly on compact subsets of D(0,1) (the unit disc centered at the origin), does is suffice to show that the sum converges uniformly on an arbitrary closed subdisc?

Subsequently, I must find the power series coefficients. How do I do this? Does anyone have general guidelines or a worked example I could analogize?

Muchas gracias
• Apr 21st 2010, 12:58 PM
Laurent
Quote:

Originally Posted by mylestone
To show that $\sum_n^{\infty} \frac{z^n}{1-z^n}$ converges uniformly on compact subsets of D(0,1) (the unit disc centered at the origin), does is suffice to show that the sum converges uniformly on an arbitrary closed subdisc?

Yes, because any compact of D(0,1) is included in such a subdisc centered at 0 (distance to the boundary is positive).

Quote:

Subsequently, I must find the power series coefficients. How do I do this? Does anyone have general guidelines or a worked example I could analogize?
The idea is to expand the ratio into a power series ( $\frac{1}{1-a}=\sum_{k=0}^\infty a^k$) so as to get a series of powers $z^m$ (for some $m$ depending on $k,n$), and gather the terms with same power $m$ by re-organizing the summation to have the usual $\sum_{p=0}^\infty a_p z^p$...
• Apr 23rd 2010, 07:55 AM
mylestone
So I have this prior result:

If for each n, $\sum_k a_k ^{(n)} z^k$ converges on a disc having radius r and being centered at 0,

and $\sum_n \sum_k a_k ^{(n)} z^k$ converges uniformly on compact subsets of the disc,

THEN, if we rewrite this as $\sum_j b_j z^j$, the coefficients $b_j$ are given by $\sum_n a_j ^{(n)}$.

It seems I should apply this but am unsure how to do so explicitly. For example, how do I write $\sum \frac{z^n}{1-z^n}$ as a double-series to apply the result?
• Apr 23rd 2010, 08:56 AM
Laurent
Quote:

Originally Posted by mylestone
It seems I should apply this but am unsure how to do so explicitly. For example, how do I write $\sum \frac{z^n}{1-z^n}$ as a double-series to apply the result?

You should definitely apply this result. As for the expression as a double-series, did you do what my last post said about expanding $\frac{1}{1-z^n}$ in power series? If not, you should...