# Complex Series Convergence

• June 5th 2011, 06:40 AM
worc3247
Complex Series Convergence
Let $\sum_{n=1}^\infty z_n$ be a series of complex numbers.. Show that if $\lim_{n \to \infty}|\frac{z_{n+1}}{z_n}|=L<1$ then there exists N such that for all positive integers k: $|z_{N+k}|\leq|z_N|(\frac{L+1}{2})^k$.

My first thoughts were to find a specific epsilon in terms of L and N to make the equation, but I couldn't find any. Could someone point me in the right direction?
• June 5th 2011, 06:49 AM
girdav
Put $\varepsilon =\frac{1-L}2$.
• June 5th 2011, 07:03 AM
worc3247
So I then get $\forall \epsilon >0 \exists N s.t. \forall n \geq N: |\frac{z_{n+1}}{z_n}-L|<\frac{1-L}{2}$. Can I then say from this that:
$|\frac{z_{n+1}}{z_n}|<\frac{1+L}{2}$?
• June 5th 2011, 08:03 AM
girdav
Since we have $\frac{z_{n+1}}{z_n}-L<\frac{1-L}2$, the answer is yes.