Analysis Divergent Sequences

  • #1

Homework Statement



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

Homework Equations



Diverges definition

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?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



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

Homework Equations



Diverges definition

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?

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:
  • #3
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
LCKurtz
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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?

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.

[Edit]: Edited it twice. I had it right the first time. :frown:
 
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  • #5
HallsofIvy
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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
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
LCKurtz
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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.

[Edit]: Edited it twice. I had it right the first time. :frown:

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}=+∞.

Did you try what I suggested? If so, how far have you gotten? Show us.
 
  • #8
I get (-n^3/2). Then I"m not sure how to solve for the N that I need.
 
  • #9
LCKurtz
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I get (-n^3/2). Then I"m not sure how to solve for the N that I need.

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
I'm sorry, I meant (-n^3)/2. So do I set (-n^3)/2 > M and solve for n?
 
  • #11
LCKurtz
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I'm sorry, I meant (-n^3)/2. So do I set (-n^3)/2 > M and solve for n?

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
N>(2M)^1/3

Is that correct?
 
  • #13
LCKurtz
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N>(2M)^1/3

Is that correct?

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.
 

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