Results 1 to 6 of 6

Math Help - Series Sum Question

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    128

    Series Sum Question

    \sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4n\sqrt{n}+4}

    \sum_{n=1}^\infty \frac{1}{\sqrt{n^3}e^n}

    The problem is to decide if the series converge or diverge. I have tried the n-th term and integral tests but to no avail.

    For the second problem, I have shown that the n-th term test does not diverge, but that does not nececarily show that it DOES converge, I believe.
    Last edited by matt.qmar; October 11th 2009 at 02:22 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1485
    Awards
    1
    On the first one it helps to realize n\sqrt{n}=\sqrt{n^3}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    128
    ok, so...
    <br />
\sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4\sqrt{n^3}+4}<br />

    Can this help to integrate? I don't see how to use the hint, sorry! Maybe I am forgetting a rule for radicals that I could use, I will take another look. Thanks though!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by matt.qmar View Post
    ok, so...
    <br />
\sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4\sqrt{n^3}+4}<br />

    Can this help to integrate? I don't see how to use the hint, sorry! Maybe I am forgetting a rule for radicals that I could use, I will take another look. Thanks though!
    it allows one to "see" that the nth term approaches \frac{\sqrt{2}}{4} as n \to \infty
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by matt.qmar View Post
    \sum_{n=1}^\infty \frac{\sqrt{2n^3+n^2-6}}{4n\sqrt{n}+4}

    \sum_{n=1}^\infty \frac{1}{\sqrt{n^3}e^n}

    The problem is to decide if the series converge or diverge. I have tried the n-th term and integral tests but to no avail.

    For the second problem, I have shown that the n-th term test does not diverge, but that does not nececarily show that it DOES converge, I believe.

    The general term of the first series doesn't even converge to zero so the series is divergent
    ]
    For the second one you ahve 1/[n^(3/2)*e^n] <= 1/n^(3/2) and apply the comparison test.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    Posts
    128
    Thank you! I forgot the simple radical division rule. Much appriciated!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. question about answer to fourier series question
    Posted in the Advanced Math Topics Forum
    Replies: 11
    Last Post: February 9th 2011, 12:29 PM
  2. Replies: 0
    Last Post: January 26th 2010, 08:06 AM
  3. Series Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 28th 2008, 12:44 AM
  4. Sequences and Series - Power Series Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2008, 07:32 PM
  5. Replies: 6
    Last Post: August 18th 2007, 12:13 PM

Search Tags


/mathhelpforum @mathhelpforum