Complex Series Convergence

**worc3247** · June 5th 2011, 06:40 AM

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?

**girdav** · June 5th 2011, 06:49 AM

Put $\varepsilon =\frac{1-L}2$ .

**worc3247** · June 5th 2011, 07:03 AM

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:
$|\frac{z_{n+1}}{z_n}|<\frac{1+L}{2}$ ?

**girdav** · June 5th 2011, 08:03 AM

Since we have $\frac{z_{n+1}}{z_n}-L<\frac{1-L}2$ , the answer is yes.

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