lamijathepanda
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I have to test the convergence of the following series. I tested the series for n=1, 2, 3, 4 and 5, and obviously the series is oscillating, but I don't know how to conclude that the series is divergent. Any help would work. Thanks.

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mqb2766
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(Original post by lamijathepanda)
I have to test the convergence of the following series. I tested the series for n=1, 2, 3, 4 and 5, and obviously the series is oscillating, but I don't know how to conclude that the series is divergent. Any help would work. Thanks.

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What is the test for convergence?
Could you find a consequence which is obviously divergent?
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lamijathepanda
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(Original post by mqb2766)
What is the test for convergence?
Could you find a consequence which is obviously divergent?
The series shall be monotone, that is the catch. This one obviously isn't because four values are repeating.
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mqb2766
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(Original post by lamijathepanda)
The series shall be monotone, that is the catch. This one obviously isn't because four values are repeating.
It's easy to pick out a subsequence of terms = 2. Adding them together ...
Equivalently group the terms so that each group sum is constant. Then sum the groups.
Note monotone isn't enough.
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lamijathepanda
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(Original post by mqb2766)
It's easy to pick out a subsequence of terms = 2. Adding them together ...
Note monotone isn't enough.
So I should find the limit?
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mqb2766
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(Original post by lamijathepanda)
So I should find the limit?
Is there one?
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lamijathepanda
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(Original post by mqb2766)
Is there one?
As far as I am concerned, no. Both values converge to infinity.
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mqb2766
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(Original post by lamijathepanda)
As far as I am concerned, no. Both values converge to infinity.
Isn't that what you call divergent? There is no finite limit.
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DFranklin
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(Original post by lamijathepanda)
I have to test the convergence of the following series. I tested the series for n=1, 2, 3, 4 and 5, and obviously the series is oscillating, but I don't know how to conclude that the series is divergent. Any help would work. Thanks.

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Can we please be clear here: are you saying the the sequence a_n = 1 + \sin \frac{n\pi}{6} diverges (that is, a_n does not tend to a limit), or are you saying that the series \sum a_n diverges? The correct approach depends on which you actually mean.
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RDKGames
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(Original post by lamijathepanda)
I have to test the convergence of the following series. I tested the series for n=1, 2, 3, 4 and 5, and obviously the series is oscillating, but I don't know how to conclude that the series is divergent. Any help would work. Thanks.

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The series converges implies a_n tends to zero in the limit n\to \infty.

It is sufficient to show this does not hold.

If a sequence has a limit L, then every subsequence has the same limit.

If we assume a_n \to 0 then we can find a subsequence which does NOT tend to 0.
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lamijathepanda
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(Original post by DFranklin)
Can we please be clear here: are you saying the the sequence a_n = 1 + \sin \frac{n\pi}{6} diverges (that is, a_n does not tend to a limit), or are you saying that the series \sum a_n diverges? The correct approach depends on which you actually mean.
The series should be tested whether they are convergent or not.
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DFranklin
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(Original post by lamijathepanda)
The series should be tested whether they are convergent or not.
Did you actually read what I posted? Because you haven't answered my question and your response makes me think you didn't even understand it.
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lamijathepanda
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(Original post by DFranklin)
Did you actually read what I posted? Because you haven't answered my question and your response makes me think you didn't even understand it.
I'm sorry about the misunderstanding. Yes, I've read what you'd posted and the series (second case you mentioned) shall be tested. Thank you for your effort to help me, I really appreciate it.
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(Original post by lamijathepanda)
I'm sorry about the misunderstanding. Yes, I've read what you'd posted and the series (second case you mentioned) shall be tested. Thank you for your effort to help me, I really appreciate it.
No problem (and I'm sorry I was a bit exasperated). In that case it's sufficient to show that a_n \not \to 0, and in this case the (arguably) easiest way to show that is find a subsequence of a_n that is constant and non-zero.
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