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

In summary: Fourth, you say "the other says diverges" when you should have said "the other says converges." Fifth, you say "I don't get it" but you never explain what you don't get. Sixth, you say "I plug ∞ in" when you should have said "I take the limit as n→∞." Seventh, you put an arrow in lim n->∞ instead of →. Eighth, you say "plug in" when you should have said "substitute." Ninth, you say "it diverges like 1/n" when you should have said "it diverges in the same way as 1/n." Tenth, you say "I mean
  • #1
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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  • #2
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.
 
  • #3
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##.
 
  • #4
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.
 
  • #5
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.
 

1. What is a convergence test?

A convergence test is a mathematical method used to determine whether a series or sequence of numbers will converge or diverge.

2. How do I know which convergence test to use?

There are several different convergence tests, such as the ratio test, the root test, and the comparison test. The specific test to use depends on the characteristics of the series or sequence being analyzed.

3. What does it mean if a series converges?

If a series converges, it means that the sum of its terms approaches a finite value as the number of terms increases. In other words, the series will eventually "settle" on a specific value.

4. What is the difference between absolute convergence and conditional convergence?

A series that converges absolutely means that the sum of the absolute values of its terms converges. In contrast, a series that converges conditionally means that the sum of its terms converges, but not the sum of the absolute values of its terms.

5. Can a series converge and diverge at the same time?

No, a series can only either converge or diverge. However, some series may require further analysis to determine whether they converge absolutely or conditionally.

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