Does the nth Term Divergence Test Conflict with the Limit Comparison Test?

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SUMMARY

The discussion centers on the apparent conflict between the nth Term Divergence Test and the Limit Comparison Test when analyzing the series represented by the expression (n^2 - 1) / (n^3 + 6n). The nth Term Divergence Test indicates convergence as the limit approaches zero, while the Limit Comparison Test suggests divergence by comparing it to the divergent series 1/n. Participants clarify that the nth term approaching zero does not guarantee series convergence, emphasizing the importance of correctly identifying whether the expression represents a sequence or a series.

PREREQUISITES
  • Understanding of the nth Term Divergence Test
  • Familiarity with the Limit Comparison Test
  • Knowledge of series and sequences in calculus
  • Ability to compute limits as n approaches infinity
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Banana Pie
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n^2 - 1 / (n^3 + 6n)

If I use the nth divergence test, I plug ∞ in (limit as n -> ∞) for n and since the degree on the bottom is larger I get 0, which means it converges.

However, if I use the limit comparison test and compare it to: n^2/n^3, which = 1/n, which diverges -> n^2 - 1 / (n^3 + 6n) / (1/n) = (n^2)(n) / (n^3 + 6) then take lim n->∞ and plug in -> =1, which is above 0, so it diverges like 1/n.

I don't get it. One method says it converges, the other says diverges. Which one do I use? Am I making a mistake? Please help me!
 
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Banana Pie said:
which means it converges.

No, taking the limit can only show that the series diverges, not that it converges. In other words, if the limit is not 0, then the series diverges, but if the limit is 0, it doesn't necessarily mean that the limit converges.
 
Banana Pie said:
n^2 - 1 / (n^3 + 6n)

Is the given expression a term of a sequence or a term of a series? It matters.

If I use the nth divergence test, I plug ∞ in (limit as n -> ∞) for n and since the degree on the bottom is larger I get 0, which means it converges.

If you mean by "it" the series for which that is the nth term that is false. The nth term going to zero does not imply covergence of the corresponding series. Although the sequence converges to zero.

However, if I use the limit comparison test and compare it to: n^2/n^3, which = 1/n, which diverges -> n^2 - 1 / (n^3 + 6n) / (1/n) = (n^2)(n) / (n^3 + 6) then take lim n->∞ and plug in -> =1, which is above 0, so it diverges like 1/n.

You mean like the series ##\sum \frac 1 n##. The sequence ##\frac 1 n## converges to ##0##.
 
Banana Pie said:
n^2 - 1 / (n^3 + 6n)
What you wrote is ##n^2 - \frac 1 {n^3 + 6n}##. Is that what you meant? If not, whenever the numerator or denominator consists of two or more terms, you need to put parentheses around the entire expression.
 
You really made a shambles of this. First, you posted homework without using the homework template. Second, you ask about "convergence" without saying if this is convergence of a sequence or a series. Third you use a test for divergence as if it were a test for convergence.
 

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