Does This Sequence Converge or Diverge?

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Homework Help Overview

The discussion revolves around determining whether a given sequence converges or diverges, with participants exploring various comparison tests and limits in the context of sequences and series.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using the Limit Comparison Test, with one suggesting a comparison with the sequence (2/3)^n. There are questions about the implications of the limit results and the validity of the tests being applied.

Discussion Status

Some participants have provided guidance on alternative comparison sequences and have noted the results of their limit comparisons. There is an ongoing exploration of the implications of these comparisons, but no explicit consensus has been reached regarding the convergence or divergence of the original sequence.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the depth of exploration or the types of methods that can be employed. There is also a focus on ensuring that comparisons are valid and correctly interpreted.

ironman
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Homework Statement


Show if this sequence (with n=1 to infinity) diverge or converge

Homework Equations


[/B]
CodeCogsEqn-5.gif


The Attempt at a Solution



If I use the Limit Comparison Test:

compare with
CodeCogsEqn-6.gif
so you get
CodeCogsEqn-7.gif
that equals
CodeCogsEqn-8.gif
lim n -> inf
CodeCogsEqn-8.gif
=> inf.

Can I use the Test like this? What does this tell me? (If anything at all...)
 
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I suggest doing limit comparison with ##(2/3)^n## instead.
 
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Orodruin said:
I suggest doing limit comparison with ##(2/3)^n## instead.

Ah the limit is now 1 .
So:
Limit < infinity.
And ##(2/3)^n## converges.
So the original function must converge as well.
 
ironman said:
Ah the limit is now 1 .
So:
Limit < infinity.
And ##(2/3)^n## converges.
So the original function must converge as well.

Right, and a direct comparison is easier. First, for ##a > 1## the function ##a^n## always swamps ##n^k## for any fixed power ##k##, meaning that for large enough ##n## we always have ##a^n > n^k##. So, if
t_n = \frac{2^n + n^2}{3^n + n^3}
we have that the numerator is ##2^n + n^2 < 2^n + 2^n = 2 \cdot 2^n## for all ##n > n_0## (where you can even find ##n_0## if you want). The denominator is ##3^n + n^3 > 3^n##, so ##0 < t_n < \frac{2 \cdot 2^n}{3^n} = 2(2/3)^n##, and now you are almost done.
 
Last edited:

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