Results 1 to 4 of 4

Math Help - testing for convergence/divergence

  1. #1
    Member
    Joined
    May 2008
    Posts
    77

    testing for convergence/divergence

    I've got a few problems here I'm wanting some guidance over.
    1. \sum_{n=0}^{\infty} \frac{n!}{2^{n^2}}
    Am I right in thinking the Ratio test is the one to use here?
    I get
    \frac {n+1}{2^{2n+1}}
    Do I use the Ratio Test again? I end up with \frac {1}{2}

    2. \sum_{n=1}^{\infty} \frac{2^n +1}{3^n -cosn}
    Absolute convergence test?
    |cos n|<1, thus the above series can be reduced down to  (\frac {2}{3})^n = \frac{2}{3}
    >>edit: sorry, made a mistake in the original post in the series equation. Now rectified.<<

    3. \sum_{n=1}^{\infty} (-1)^{n-1(n^\frac{1}{n}-1)^n}

    I'm thinking the Root test here?
    then, since \lim_{n \to\infty} {n}^\frac{1}{n} = 1, we're left with just (-1)^n

    Any/all of these on the right track or have I totally lost the plot here?
    Cheers!
    Last edited by Dr Zoidburg; March 23rd 2010 at 03:33 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,695
    Thanks
    1510
    Quote Originally Posted by Dr Zoidburg View Post
    I've got a few problems here I'm wanting some guidance over.
    1. \sum_{n=0}^{\infty} \frac{n!}{2^{n^2}}
    Am I right in thinking the Ratio test is the one to use here?
    I get
    \frac {n+1}{2^{2n+1}}
    Do I use the Ratio Test again? I end up with \frac {1}{2}

    2. \sum_{n=1}^{\infty} \frac{2n+1}{3-cosn}
    Absolute convergence test?
    |cos n|<1, thus the above series can be reduced down to  (\frac {2}{3})^n = \frac{2}{3}

    3. \sum_{n=1}^{\infty} (-1)^{n-1(n^\frac{1}{n}-1)^n}

    I'm thinking the Root test here?
    then, since \lim_{n \to\infty} {n}^\frac{1}{n} = 1, we're left with just (-1)^n

    Any/all of these on the right track or have I totally lost the plot here?
    Cheers!
    1. \sum_{n=0}^{\infty} \frac{n!}{2^{n^2}}

    Yes, use the ratio test.

    You need to find

    \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|

     = \lim_{n \to \infty}\left|\frac{\frac{(n + 1)!}{2^{(n + 1)^2}}}{\frac{n!}{2^{n^2}}}\right|

     = \lim_{n \to \infty}\left|\frac{(n + 1)!2^{n^2}}{n!2^{(n + 1)^2}}\right|

     = \lim_{n \to \infty}\left|\frac{(n + 1)2^{n^2}}{2^{n^2 + 2n + 1}}\right|

     = \lim_{n \to \infty}\left|\frac{n + 1}{2^{2n + 1}}\right|

     = \lim_{n \to \infty}\frac{n + 1}{2^{2n + 1}} since it's a positive term series

     = \lim_{n \to \infty}\frac{1}{2\ln{2}\cdot 2^{2n + 1}} by L'Hospital's Rule

     = 0.


    Since this limit < 0, the series is convergent.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    May 2008
    Posts
    77
    thanks for that.
    Out of curiosity, can I just do a Ratio test again instead of use L'Hopital's Rule?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,695
    Thanks
    1510
    Quote Originally Posted by Dr Zoidburg View Post
    thanks for that.
    Out of curiosity, can I just do a Ratio test again instead of use L'Hopital's Rule?
    No, because you don't get another series. You get a term that you're finding the limit of...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Testing Series for Convergence/Divergence
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 16th 2010, 03:39 PM
  2. Testing series for convergence or divergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 27th 2010, 08:33 PM
  3. Testing a series for convergence or divergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 22nd 2010, 10:18 PM
  4. Testing for Convergence/Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 07:08 PM
  5. Testing for Convergence/Divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 06:19 PM

Search Tags


/mathhelpforum @mathhelpforum