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 Complex Analysis Complex Analysis Math Forum

 January 26th, 2018, 03:55 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 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: 7,690 Thanks: 2669 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: 884 Thanks: 340 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 ||, $\displaystyle |z_{n+1}|\lt \alpha |z_n|$ Tags absolute, complex, convergence, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post joesmith Real Analysis 0 November 29th, 2014 01:29 PM Chemist@ Calculus 7 November 1st, 2014 02:16 PM linasasuk Complex Analysis 2 April 6th, 2012 06:46 AM David_Lete Calculus 2 January 31st, 2011 07:13 PM draxler Calculus 1 November 6th, 2008 12:41 PM

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