# Analysis Divergent Sequences

1. Sep 21, 2012

### MathSquareRoo

1. The problem statement, all variables and given/known data

Prove that the given sequence diverges to infinity.
{an} = (-n^4+n^3+n)/(2n+7)

2. Relevant equations

Diverges definition

3. The attempt at a solution

So far I have:

Let M>0 and let N= something.

I'm having a hard time figuring out what N should equal for the proof. I know I have to make the numerator smaller and denominator larger (I think), but I get confused. Can someone explain it please?

2. Sep 21, 2012

### LCKurtz

The first thing you must notice is with that minus sign in the numerator on the $n^4$ term, this isn't going to go to $+\infty$. It will go to $-\infty$. Let's call your fraction $f(n)$ for the moment. Start by writing the careful definition of the statement$$\lim_{n\rightarrow \infty}f(n) = -\infty$$Then you will know what you must prove.

[Edit, added later]: Alternatively you could show $|f(n)|\rightarrow \infty$ and observe that $f(n)<0$ for large $n$.

Last edited: Sep 21, 2012
3. Sep 21, 2012

### MathSquareRoo

Yes, I understand that it will diverge to negative infinity. I still need help understanding how to get the N for the proof. Can you help?

4. Sep 21, 2012

### LCKurtz

I don't know if you saw may later edit, which is the method I would use. Given $M>0$, to find $N$ you generally use an exploratory argument which is kind of working backwards with estimates. So start with$$\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...$$and underestimate it to get something workable. I would start with the old calculus trick of dividing numerator and denominator by the highest power of $n$. Then work on making the denominator larger and the numerator smaller to underestimate it with something that still goes to infinity.

: Edited it twice. I had it right the first time.

Last edited: Sep 21, 2012
5. Sep 22, 2012

### HallsofIvy

Staff Emeritus
For sufficiently large n, the highest degree term will dominate. For sufficiently large n, (-n^4+n^3+n)/(2n+7) will be approximately -(n^3)/2.

6. Sep 23, 2012

### MathSquareRoo

I'm sorry, I still don't understand how to write the proof for this divergent sequence. I have a similar problem done correctly I believe. Can someone help me write a proof like the following one, but for this sequence above??

Here is a similar lim, I think I have correctly proved.

{an}=(n^2+1)/(n-2)

Proof: Let M>0 and let N=2M. Then n>N implies 1/2n>M which implies (n^2+1)/(n-2)<n^2/2n=1/2n>M. Hence, lim n--> {an}=+∞.

7. Sep 23, 2012

### LCKurtz

Did you try what I suggested? If so, how far have you gotten? Show us.

8. Sep 23, 2012

### MathSquareRoo

I get (-n^3/2). Then I"m not sure how to solve for the N that I need.

9. Sep 23, 2012

### LCKurtz

I have no idea where you got that but, as an example of solving for $N$, you are given an $M$ and you want $n^{3/2} > M$ for $n > N$. So solve the inequality $n^{3/2} > M$ for $n$ to see how large $n$ needs to be. Then choose $N$ so that if $n>N$ you can guarantee $n$ is as large as it needs to be to make it work.

10. Sep 23, 2012

### MathSquareRoo

I'm sorry, I meant (-n^3)/2. So do I set (-n^3)/2 > M and solve for n?

11. Sep 23, 2012

### LCKurtz

Yes, same idea. You want the absolute value though: $\frac{n^3}{2}>M$. So how big does $n$ need to be? What $N$ will work?

12. Sep 23, 2012

### MathSquareRoo

N>(2M)^1/3

Is that correct?

13. Sep 24, 2012

### LCKurtz

That is how you determine how large N needs to be with an exploratory argument. Now, at this point, we have, if $n>N$ $$\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...\hbox{ your steps here }...\ge n^{\frac 3 2}> M$$

To write up the argument you would phrase it like this: Given $M>0$ if $n>(2M)^\frac 1 3$ then$$\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...\hbox{ your steps here }...\ge n^{\frac 3 2}> M$$Notice in this last paragraph, it looks like you magically came up with $n>(2M)^\frac 1 3$, but it came from the exploratory argument.

I hope this has helped you understand the argument. Also note that since you never posted how you got the $\frac {n^3} 2$, I am not vouching for your complete argument. I will leave that to you.