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January 26th, 2018, 04:55 AM   #1
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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.
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January 26th, 2018, 07:31 AM   #2
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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.
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January 26th, 2018, 09:00 AM   #3
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$z_{n+1} < \alpha z_n$ makes no sense in the complex plane.
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January 26th, 2018, 10:01 AM   #4
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Not canonically at least.

But I'm sure the OP can fix this small issue.
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January 26th, 2018, 01:29 PM   #5
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Originally Posted by Micrm@ss View Post
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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