0
$\begingroup$

I want to prove that if $(a_n)$ is a complex sequence such that $\sum a_n = 0 $ then I can define an holomorphic function $f(z)$ on $ D:= \{z\ | \ |z|<1\} $ by $f(z)= \sum a_n z^n$

In order to do this I wanted to consider the subsequence $(a_{n_k})$ of the nonzero terms of $(a_n)$ and hence use a converse of the ratio test, i.e. say that $\lim_{k \rightarrow \infty}{|a_{n_{k+1}}/a_{n_k}}|\leq 1$

Hence it would be easy to prove that $f$ actually is holomorphic in $D$

My concern is that $\lim_{k \rightarrow \infty}{|a_{n_{k+1}}/a_{n_k}}|$ might not be defined (even if the terms are all non zero..) any suggestion on how to prove it rigorously?

1 Answers 1

3

The ratio test doesn't really have a converse. But all you need in this case is that if the sequence $(a_n)$ of coefficients is bounded, then $\sum a_nz^n$ converges for $\lvert z\rvert<1$. You prove that via a direct comparison with the convergent geometric series $\sum \lvert z\rvert^n$.

  • 0
    you're right, thank you! i overcomplicated it a bit!2012-11-09