So I have this complex sequence:

$$f(z)\frac{z-19}{z^2-8z-9}= \sum_{k=0}^{\infty} \left(\left(\frac{1}{9}\right)^{k+1}+2\left(-1\right)^k\right)z^k$$

I am trying to show that the radius of convergence is $1$ using a convergence test to find $L$. Anyone have any idea? I trying using the ratio test $lim_{n \rightarrow \infty}=\dfrac{z_{n+1}}{z_n}$ But cannot seem to show this. If anyone has any idea how to simply do this it would be greatly appreciated.


1 Answer 1


We need $$\dfrac19\sum_{k=0}^\infty\left(\dfrac z9\right)^k+2\sum_{k=0}^\infty(-z)^k$$

Now use this


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