Results 1 to 6 of 6

Math Help - Determine the convergence.

  1. #1
    Junior Member
    Joined
    Mar 2008
    Posts
    49

    Determine the convergence.

    For the following series ∑ of a(n) between n=1 and ∞, determine if
    the series converges and if so find the limit. (Hint: use partial
    fractions to express a(n).)

    (a) a(n)=1/(n(n+1))

    (b) a(n)=1/(n^2+2n)

    (c) a(n)=1/(n(n^2-1))
    (As a(1) is not defined consider ∑ of a(n) between n=2 and ∞.)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    #1:

    Are you familiar with telescoping sums?.

    \sum_{1}^{\infty}\frac{1}{n(n+1)}

    =\sum_{1}^{\infty}\left[\frac{1}{n}-\frac{1}{n+1}\right]

    =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+......

    Notice what is going on?.

    Everything cancels one another and you're left with the 1 in the very beginning. So, it converges to 1.

    The others may be similar. Check and see.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Quote Originally Posted by galactus View Post
    \sum_{1}^{\infty}\frac{1}{n(n+1)}
    Just for fun, I just solve this one as follows:

    \sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n(n + 1)}}}  = \int_0^1 {\left\{ {\sum\limits_{n = 1}^\infty  {\frac{{x^n }}<br />
{n}} } \right\}\,dx}  =  - \int_0^1 {\ln (1 - x)\,dx}  = \int_0^1 \!{\int_0^{\,x} {\frac{{dy\,dx}}<br />
{{1 - y}}} } .

    So

    \sum\limits_{n = 1}^\infty  {\frac{1}<br />
{{n(n + 1)}}}  = \int_0^1 \!{\underbrace {\int_y^1 {dx} }_{1\, -\, y}\,\frac{{dy}}<br />
{{1 - y}}}  = 1,

    as required.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    Ok i did part b and got it converges to 3/4

    because  1/(n^2+2n)= (1/2n)-(1/2n+4)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    c.
    You can compare with \zeta(s)=\sum_{n=1}^{\infty}{\frac{1}{n^s}}

    This series converges if s>1

    We have (n>1): n^2-1<n^2 ==> n\cdot{(n^2-1)}<n^3 so \frac{1}{n^3}<\frac{1}{n\cdot{(n^2-1)}}

    Thus it follows that the series converges by comparison with \zeta(3)



    To calculate the sum:
    \frac{1}{n\cdot{(n^2-1)}}=\left(\frac{1}{n+1}\right)\cdot{\left(\frac{1  }{n-1}-\frac{1}{n}\right)}

    So; \frac{1}{n\cdot{(n^2-1)}}=\frac{1}{(n+1)\cdot(n-1)}-\frac{1}{(n+1)\cdot(n)}

    Summing: \sum_{n=2}^{\infty}\frac{1}{n\cdot{(n^2-1)}}=\sum_{n=2}^{\infty}\frac{1}{(n+1)\cdot(n-1)}-\sum_{n=2}^{\infty}\frac{1}{(n+1)\cdot{n}}

    Both series on the right are telescoping series

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2008
    Posts
    49
    Ok i got

    a) 1

    b) 3/4

    c) 1/4

    can any1 tell me if there right.

    And is there in index of how to use the math tags for this forum [tex] those.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. determine the convergence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 13th 2011, 02:52 PM
  2. Determine convergence or divergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 17th 2010, 07:46 PM
  3. How to determine the convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 6th 2010, 03:46 PM
  4. determine interval of convergence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 13th 2009, 12:37 AM
  5. Determine the Interval of Convergence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 13th 2008, 02:01 PM

Search Tags


/mathhelpforum @mathhelpforum