How do you test the convergence of i^n (imaginary number) or (-1)^n?

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  • #1
Agnostic
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I was thinking alternating series test, but can't that only be used for testing for convergence, not for testing divergence?
 

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  • #2
Hurkyl
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How about the divergence test? :tongue2:
 
  • #3
LeonhardEuler
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You'll have to be more specific. The convergence of what? The sequence of terms, [itex] a_n=i^n[/itex], the series, [itex]\sum_{n=0}^{\infty}i^n[/itex], or some sequence or series that just has [itex]i^n[/itex] as one term? I can tell you right now, the first two diverge.
 
  • #4
Agnostic
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Hurkyl said:
How about the divergence test? :tongue2:

The divergence test doesn't work for indeterminant forms.
 
  • #5
Agnostic
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LeonhardEuler said:
You'll have to be more specific. The convergence of what? The sequence of terms, [itex] a_n=i^n[/itex], the series, [itex]\sum_{n=0}^{\infty}i^n[/itex], or some sequence or series that just has [itex]i^n[/itex] as one term? I can tell you right now, the first two diverge.


I must prove that [itex]\sum_{n=0}^{\infty}i^n[/itex] diverges.

Or rather, I must prove wheter or not its convergent of divergent. I know its divergent, but I still have to prove it.
 
  • #6
LeonhardEuler
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Agnostic said:
The divergence test doesn't work for indeterminant forms.
[itex]i^n[/itex] is not an indeterminant form.
 
  • #7
Hurkyl
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The divergence test doesn't work for indeterminant forms.

There isn't an indeterminant form involved. :tongue2:
 
  • #8
LeonhardEuler
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Agnostic said:
I must prove that [itex]\sum_{n=0}^{\infty}i^n[/itex] diverges.

Or rather, I must prove wheter or not its convergent of divergent. I know its divergent, but I still have to prove it.
Oh, that's simple. Use the divergence test as Hurkyl suggested. Do the terms in the series approach zero as n approaches infinty?
 
  • #9
Agnostic
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The limit of [itex]i^n[/itex] as n goes to infinity doesn't have a solution does it?
 
  • #10
LeonhardEuler
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Agnostic said:
The limit of [itex]i^n[/itex] as n goes to infinity doesn't have a solution does it?
Right, exactly!
 
  • #11
Agnostic
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LeonhardEuler said:
Oh, that's simple. Use the divergence test as Hurkyl suggested. Do the terms in the series approach zero as n approaches infinty?

No, they alternate ... i,-1,-i,1,i,-1,...
 
  • #12
Agnostic
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LeonhardEuler said:
Right, exactly!

But that is not the same as saying indeterminant form?
 
  • #13
LeonhardEuler
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Agnostic said:
No, they alternate ... i,-1,-i,1,i,-1,...
So the limit doesn't exist. Then can it equal zero?
 
  • #14
LeonhardEuler
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Agnostic said:
But that is not the same as saying indeterminant form?
No, an indeterminant form is something like [itex]\frac{0}{0}[/itex], or [itex]\frac{\infty}{\infty}[/itex], or [itex]1^{\infty}[/itex]. A non-existant limit is just a non-existant limit.
 
  • #15
Hurkyl
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But that is not the same as saying indeterminant form?

An indeterminate form is a "limit form" for which we don't have enough information to say what the limit really is, or if it exists. For example, [itex]\lim_{x \rightarrow 0} x/x[/itex] has the indeterminate form 0/0.
 
  • #16
Agnostic
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Its usually always silly mistakes :D
Thanks a lot guys..
 

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