# VERY CONFUSED with convergence test?

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1. Mar 16, 2016

### Banana Pie

• Member warned about posting without the template
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!

2. Mar 16, 2016

### axmls

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. Mar 16, 2016

### LCKurtz

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

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.

You mean like the series $\sum \frac 1 n$. The sequence $\frac 1 n$ converges to $0$.

4. Mar 16, 2016

### Staff: Mentor

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. Mar 17, 2016

### HallsofIvy

Staff Emeritus
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.