Results 1 to 7 of 7

Math Help - My method of proving divergence

  1. #1
    Senior Member sfspitfire23's Avatar
    Joined
    Oct 2009
    Posts
    273

    My method of proving divergence

    Prove: \frac{n^2+1}{n-2} diverges


    For every L there exists an epsilon greater than zero, for every N greater than 0, n>0.

    So, |\frac{n^2+1}{n-2}-L|>\epsilon

    Pf: Let L be given. Set \epsilon=.001 and let N>0 be given. There exists n_0>N s.t. n_0=3. Then, if |\frac{n_0^2+1}{n_0-2}-L|<\epsilon, we would have |10-L|<\epsilon \implies -.001<L-10<.001 \implies 9.999<L<10.001. But then if |\frac{n_{0+1}^2+1}{n_{0+1}-2}-L|<\epsilon we would have |8.5-L|<\epsilon. So, 8.499<L<8.501. But then L would be in different intervals. Thus the sequence diverges as desired.

    Is this the correct way to go about this?

    Thx!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by sfspitfire23 View Post
    Prove: \frac{n^2+1}{n-2} diverges


    For every L there exists an epsilon greater than zero, for every N greater than 0, n>0.

    So, |\frac{n^2+1}{n-2}-L|>\epsilon

    Pf: Let L be given. Set \epsilon=.001 and let N>0 be given. There exists n_0>N s.t. n_0=3. Then, if |\frac{n_0^2+1}{n_0-2}-L|<\epsilon, we would have |10-L|<\epsilon \implies -.001<L-10<.001 \implies 9.999<L<10.001. But then if |\frac{n_{0+1}^2+1}{n_{0+1}-2}-L|<\epsilon we would have |8.5-L|<\epsilon. So, 8.499<L<8.501. But then L would be in different intervals. Thus the sequence diverges as desired.

    Is this the correct way to go about this?

    Thx!
    It looks as though you have proved that the sequence does not converge to any L\in\mathbb{R}. Is that what they want you to do?

    More informative would be to prove the rigorous idea that \frac{n^2+1}{n-2}\to\infty.

    To do this first note that \frac{n^2+1}{n-2}\geqslant n+\frac{1}{n}\geqslant n and so given any T>0 we may choose N\in\mathbb{N} (namely N=\left\lceil T\right\rceil) such that N\leqslant n\implies T<n. The conclusion follows.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    It doesn't need to be that complicated. Try replacing the numerator with something smaller, and the denominator with something larger. Choose your replacements in such a way that the quotient simplifies to something that obviously diverges. Then you'll know that the original sequence is larger than a sequence that diverges.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member sfspitfire23's Avatar
    Joined
    Oct 2009
    Posts
    273
    Ah ok....

    How would you recognize that
    <br /> <br />
\frac{n^2+1}{n-2}\geqslant n+\frac{1}{n}\geqslant n<br />
?

    Experience? haha
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by sfspitfire23 View Post
    Ah ok....

    How would you recognize that
    <br /> <br />
\frac{n^2+1}{n-2}\geqslant n+\frac{1}{n}\geqslant n<br />
?

    Experience? haha
    That should be pretty apparent. Making the denominator smaller makes the whole fraction larger so increasing the denominator makes the fraction smaller, etc.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member sfspitfire23's Avatar
    Joined
    Oct 2009
    Posts
    273
    So, to show divergence to infinity I can find a smaller sequence than my sequence and play with that.


    To show convergence I can show a larger sequence than my sequence converges


    Yeah?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by sfspitfire23 View Post
    So, to show divergence to infinity I can find a smaller sequence than my sequence and play with that.


    To show convergence I can show a larger sequence than my sequence converges


    Yeah?
    The first is correct - if your sequence is greater than a sequence which "goes" to infinity, then so does your sequence.

    The second conclusion is not always correct, though. Consider a_n = 1+\frac{1}{n}, ~ b_n = sin(n), then obviously b_n < a_n ~ \forall n \in \mathbb{N} and a_n \to 1 but b_n does not converge.

    It is correct, though, that a bounded, monotone sequence converges.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Method of difference (proving)
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 15th 2011, 09:59 PM
  2. Trouble proving convergence/divergence
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: January 7th 2011, 09:39 AM
  3. Element Method for proving...
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 12th 2010, 01:21 AM
  4. Proving convergence of divergence of a series.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2009, 02:14 AM
  5. Newton method : divergence
    Posted in the Calculus Forum
    Replies: 0
    Last Post: May 22nd 2008, 10:49 PM

Search Tags


/mathhelpforum @mathhelpforum