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Analysis Divergent Sequences

  1. Sep 21, 2012 #1
    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. jcsd
  3. Sep 21, 2012 #2

    LCKurtz

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    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
  4. Sep 21, 2012 #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?
     
  5. Sep 21, 2012 #4

    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:
     
    Last edited: Sep 21, 2012
  6. Sep 22, 2012 #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.
     
  7. Sep 23, 2012 #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}=+∞.
     
  8. Sep 23, 2012 #7

    LCKurtz

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    Did you try what I suggested? If so, how far have you gotten? Show us.
     
  9. Sep 23, 2012 #8
    I get (-n^3/2). Then I"m not sure how to solve for the N that I need.
     
  10. Sep 23, 2012 #9

    LCKurtz

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    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.
     
  11. Sep 23, 2012 #10
    I'm sorry, I meant (-n^3)/2. So do I set (-n^3)/2 > M and solve for n?
     
  12. Sep 23, 2012 #11

    LCKurtz

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    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?
     
  13. Sep 23, 2012 #12
    N>(2M)^1/3

    Is that correct?
     
  14. Sep 24, 2012 #13

    LCKurtz

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    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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