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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?
Hurkyl said:How about the divergence test?
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.
[itex]i^n[/itex] is not an indeterminant form.Agnostic said:The divergence test doesn't work for indeterminant forms.
The divergence test doesn't work for indeterminant forms.
Oh, that's simple. Use the divergence test as Hurkyl suggested. Do the terms in the series approach zero as n approaches infinty?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.
Right, exactly!Agnostic said:The limit of [itex]i^n[/itex] as n goes to infinity doesn't have a solution does it?
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?
LeonhardEuler said:Right, exactly!
So the limit doesn't exist. Then can it equal zero?Agnostic said:No, they alternate ... i,-1,-i,1,i,-1,...
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.Agnostic said:But that is not the same as saying indeterminant form?
But that is not the same as saying indeterminant form?
The convergence of i^n or (-1)^n can be tested using the ratio test or the root test. These tests involve taking the limit of the absolute value of the ratio or root of the n+1 term over the nth term as n approaches infinity. If the limit is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent. If the limit is equal to 1, the test is inconclusive and another test should be used.
The ratio test and the root test are both used to test the convergence of a series. The main difference between these tests is the type of limit that is taken. The ratio test involves taking the limit of the ratio of the n+1 term over the nth term, while the root test involves taking the limit of the nth root of the absolute value of the n+1 term. Both tests use the same criteria for determining convergence or divergence.
No, the comparison test cannot be used to test the convergence of i^n or (-1)^n because these series involve imaginary numbers. The comparison test is used for real-valued series only. The ratio test or the root test should be used instead.
The series of i^n or (-1)^n is absolutely convergent if the series of the absolute values of i^n or (-1)^n is convergent. It is conditionally convergent if the series of i^n or (-1)^n is convergent but the series of the absolute values of i^n or (-1)^n is divergent. The absolute convergence or conditional convergence of these series can be determined using the ratio test or the root test.
Yes, the convergence of i^n or (-1)^n can be determined using the integral test. This test involves comparing the series to an improper integral and using the comparison property of integrals to determine convergence or divergence. However, this test may not always be the most efficient method for testing convergence and other tests such as the ratio test or the root test may be more appropriate.