Find the limit of sequence: n(n-1)

[MarkSA]

Hello,

1) Find the limit of sequence: n(n-1)

I know that the answer is infinity, divergent. I'm not sure how it's arrived at though.

If multiplied through, this is n^2 - n. When it goes to infinity, that is infinity minus infinity which is indeterminate.

How would I solve such limit? Thanks

 

[stapel]

What tests do you have?

Thank you!

Eliz.

 

[sgtpepper]

I don't know if there is a test for this - there probably is, but if you don't know if, you could just try plugging in the first few number for n (I'm assuming that your first term is either 0 or 1).

For n = 0, you'd get:

0(0-1) = 0

for n = 1, you'd get:

1(1-1) = 0

for n = 2, you'd get:

2(2-1) = 2

for n = 3, you'd get:

3(3-1) = 6

for n = 4, you'd get:

4(4-1) = 12

Hopefully you can see that as n increases, the value of the function does too. This means it doesn't go to a certain value; instead, it approaches infinity, which is divergent. Kind of get it? See what the limit would be for this:

1 / (n(n-1))

you should get 0, convergent.

I think a rule you can remember with sequences is that if the function is top-heavy, like your one is, the sequence diverges. If it's bottom heavy, like the one I gave you, the limit is zero, and it converges. However if the function has like terms, then the limit is the ratio of the leading coefficients and the sequence converges.

i.e.

[(2n^2) + (5n) + 4] / [1 + (3n^2)]

because there is an n^2 in both the top and bottom, the terms are the same and so the limit of the sequence will be their ratio, or 2/3. I hope this helps somewhat and that I didn't mislead you.

 

[stapel]

I know that the answer is infinity, divergent. I'm not sure how it's arrived at though.

I don't know if there is a test for this - there probably is, but if you don't know if, you could just try plugging in the first few number for n....

You seem to be operating under the misapprehension that the poster is asking what the limit might be...?

Plugging some numbers into the expression, as you advocate, might give one an idea of the limit, but it does not prove the limit. If you study the first post in this thread, you will discover that the poster needs to prove that the limit is what he already knows it to be.

To MarkSA: Do you have any "known good" limits, such as "we know that the sequence n has no limiting value, but instead tends toward infinity"? And can you use the Comparison Test? If so, then, since n - 1 > 2 - 1 for all n > 2, you can state that n(n - 1) > n(2 - 1) = n(1) = n, so n(n - 1) is larger than n for all n > 2, etc, etc.

Eliz.

 

[MarkSA]

sgtpepper, thanks for the reply, I didn't know about those tricks. It looks obvious when you plug numbers in that it goes to infinity, but i'm not sure if i'm allowed to just say it goes to infinity.

stapel, I'm not sure what you mean by known good limits. I don't think we can use the comparison test yet, we haven't yet learned it in class.

 

[stapel]

I'm not sure what you mean by known good limits.

I meant something along the lines of the example I provided. Do you "know" (from the book or from class) that {n} "goes to infinity"? If this has already been established (proved, stated, "a given"), then you can work from that result in the manner suggested.

I don't think we can use the comparison test yet, we haven't yet learned it in class.

Okay; what tests have you learned? Or are you still working from the definitions?

If the latter, then do the definition-based proof:

If the sequence a[sub:193rsn4n]n[/sub:193rsn4n] = {n(n - 1)} has some limit L, then, for any "distance" d > 0, there is some natural number N so that |a[sub:193rsn4n]n[/sub:193rsn4n] - L| < d for all n > N. Suppose this is so. Then....

...and so forth. Is this the sort of proof you're needing to do?

Thank you!

Eliz.

 

[MarkSA]

This is the first section we've done on sequences. I think the only definition we've been given is:

A sequence a(sub n) has the limit L and we write:
lim as n -> infinity of: a(sub n) = L

And this bit in a form: |an - L| < d for all n > N. I guess that means the same as the above definition...

I've pretty much just been solving these by putting the sequence into a limit to infinity and solving the limit to determine convergent/divergent. I just realized I forgot to write the instructions they gave us to solve the problem in the initial post. Sorry about that:

"Determine whether the sequence converges or diverges. If it converges, find the limit."