Results 1 to 5 of 5

Math Help - converge or diverge

  1. #1
    Member
    Joined
    Jun 2007
    Posts
    120

    converge or diverge

    Can you guys check my work to see if I did it right

    the function is [(n^2)(2^n-1)]/(-5^n)

    so what i did was use the nth root test and got (n^(2/n))/10..(is the lim of n^(2/n) equal to 1??..i just used my calculator for this..)

    so the limit of that would be 1/10 <1 therefore converge

    am i right?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1573
    Awards
    1
    Quote Originally Posted by johntuan View Post
    the function is [(n^2)(2^n-1)]/(-5^n)
    There are some real notational problems to this question.

    -5^n is a negative number. \sum {\frac{1}{{ - 5^n }} =  - \sum {\frac{1}{{5^n }}} } .

    Do you mean (-5)^n?


    Is it 2^n-1 or 2^{n-1}?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2007
    Posts
    120
    Quote Originally Posted by Plato View Post
    There are some real notational problems to this question.

    -5^n is a negative number. \sum {\frac{1}{{ - 5^n }} =  - \sum {\frac{1}{{5^n }}} } .

    Do you mean (-5)^n?


    Is it 2^n-1 or 2^{n-1}
    sorry about the notation problem

    for the (-5), I thought it didnt matter because when using the nth-root test, its the absolute value that's using.

    so yeah i meant (-5)^n

    and 2^{n-1}? is the correct one.

    sorry for the confusion
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396
    Hi

    I assume you mean \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}

    |\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}|\leq \frac{n^{2}(2^{n})}{5^{n}}

    So this series is absolute convergent.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2007
    Posts
    120
    Quote Originally Posted by Twig View Post
    Hi

    I assume you mean \sum_{n=1}^{\infty}\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}

    |\frac{(-1)^{n}n^{2}(2^{n}-1)}{5^{n}}|\leq \frac{n^{2}(2^{n})}{5^{n}}

    So this series is absolute convergent.
    no theres no (-1)^n in the numerator and (-5^n) on the denominator
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Converge or diverge
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 15th 2010, 08:58 PM
  2. Converge or Diverge?
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 15th 2010, 08:39 AM
  3. Converge or diverge?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 22nd 2009, 08:54 PM
  4. Converge/Diverge help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 9th 2009, 11:57 AM
  5. Diverge/Converge
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 14th 2008, 01:55 PM

Search Tags


/mathhelpforum @mathhelpforum