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 January 26th, 2018, 03:55 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 103 Thanks: 2 absolute convergence of complex series. Let $z_{n}$ be a complex series such that $\frac{z_{n+1}}{z_{n}}$ tends to $x$ as n tends to infinity where x<1. Show $z_{n}$ converges absolutely.
January 26th, 2018, 06:31 AM   #2
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Quote:
 Originally Posted by Jaket1 Let $z_{n}$ be a complex series such that $\frac{z_{n+1}}{z_{n}}$ tends to $x$ as n tends to infinity where x<1. Show $z_{n}$ converges absolutely.
For large $n$, we have

$$z_{n+1} < \alpha z_n$$
for $x<\alpha<1$.

Now compare with a geometric series.

 January 26th, 2018, 08:00 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,636 Thanks: 2621 Math Focus: Mainly analysis and algebra $z_{n+1} < \alpha z_n$ makes no sense in the complex plane.
 January 26th, 2018, 09:01 AM #4 Senior Member   Joined: Oct 2009 Posts: 771 Thanks: 278 Not canonically at least. But I'm sure the OP can fix this small issue.
January 26th, 2018, 12:29 PM   #5
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Quote:
 Originally Posted by Micrm@ss For large $n$, we have $$z_{n+1} < \alpha z_n$$ for $x<\alpha<1$. Now compare with a geometric series.
Use ||, $\displaystyle |z_{n+1}|\lt \alpha |z_n|$

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