Results 1 to 7 of 7

Math Help - Comparison test

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Comparison test

    Use the comparison test to determine whether each of the following series converges or diverges.

    \sum \frac{2n^2+15n}{n^3+7}
    This is harder than it looks. My supervisor and I got stuck on it!

    Here's what I tried:

    0< \sum \frac{2n^2+15n}{n^3+7}< \sum \frac{2n^2+15n+7}{n^3+7}

    = \sum \frac{(2n+1)(n+7)}{n^3+7}= \sum \frac{2n(n+7)}{n^3+7}+ \frac{n+7}{n^3+7}< \sum \frac{2n^2+14n}{n^3}+\sum \frac{n+7}{n^3}

    =\sum \frac{2}{n}+\sum \frac{14}{n^2}+\sum \frac{1}{n^2} +\sum \frac{7}{n^3}

    This was the closest I got. I have to find a way to get rid of \sum \frac{2}{n} since that's the only one that diverges.

    However, i'm assuming that the sequence is convegent since I couldn't get it to diverge by the comparison test either!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by Showcase_22 View Post
    However, i'm assuming that the sequence is convegent since I couldn't get it to diverge by the comparison test either!
    This series diverges since for n\geq1

    \frac{2n^2+15n}{n^3+7}\geq \frac{2n^2+14n}{n^3+7n^2}=\frac{2}{n}.
    Last edited by flyingsquirrel; November 10th 2008 at 06:04 AM. Reason: \leq --> \geq
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Showcase_22 View Post
    This is harder than it looks. My supervisor and I got stuck on it!

    Here's what I tried:

    0< \sum \frac{2n^2+15n}{n^3+7}< \sum \frac{2n^2+15n+7}{n^3+7}

    = \sum \frac{(2n+1)(n+7)}{n^3+7}= \sum \frac{2n(n+7)}{n^3+7}+ \frac{n+7}{n^3+7}< \sum \frac{2n^2+14n}{n^3}+\sum \frac{n+7}{n^3}

    =\sum \frac{2}{n}+\sum \frac{14}{n^2}+\sum \frac{1}{n^2} +\sum \frac{7}{n^3}

    This was the closest I got. I have to find a way to get rid of \sum \frac{2}{n} since that's the only one that diverges.

    However, i'm assuming that the sequence is convegent since I couldn't get it to diverge by the comparison test either!
    Or try the limit comparison test, comparing the the Harmonic Series.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    What's the limit form of the comparison test?

    I thought it would be the same as what i'm already doing except for the addition of "lim" in front of every sum symbol.

    =S
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Showcase_22 View Post
    What's the limit form of the comparison test?

    I thought it would be the same as what i'm already doing except for the addition of "lim" in front of every sum symbol.

    =S
    It roughly states: Let \sum{a_n} and \sum{b_n} be two infinite series. Then if \lim_{n\to\infty}\frac{a_{n}}{b_n}=c\in\mathbb{R^+  }\implies\sum{a_n}\text{ and }\sum{b_n}\text{ share convergence/divergence}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782
    Quote Originally Posted by Mathstud28 View Post
    It roughly states: Let \sum{a_n} and \sum{b_n} be two infinite series. Then if \lim_{n\to\infty}\frac{a_{n}}{b_n}=c\in\mathbb{R^+  }\implies\sum{a_n}\text{ and }\sum{b_n}\text{ share convergence/divergence}
    I haven't seen this before but there's something I was wondering:

    Your \Rightarrow only goes one way so I can't use the limits of the numerator and the denominator to imply that the fraction will either converge or diverge.

    Am I applying the limit form of the comparison test to the wrong sequence?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by Showcase_22 View Post
    I haven't seen this before but there's something I was wondering:

    Your \Rightarrow only goes one way so I can't use the limits of the numerator and the denominator to imply that the fraction will either converge or diverge.

    Am I applying the limit form of the comparison test to the wrong sequence?
    If your asking that if \sum{a_n} and \sum{b_n} share convergence/divergence then this implies that \lim_{n\to\infty}\frac{a_n}{b_n}=c\in\mathbb{R^+}. If so the answer is you can, it is just not pertinent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Comparison test and Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 5
    Last Post: November 25th 2010, 12:54 AM
  2. Comparison or Limit Comparison Test Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 12th 2010, 07:46 AM
  3. Limit comparison/comparison test series
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 25th 2009, 08:27 PM
  4. Comparison & Limit Comparison test for series
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 25th 2009, 04:00 PM
  5. Integral Test/Comparison Test
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 27th 2009, 10:58 AM

Search Tags


/mathhelpforum @mathhelpforum