Results 1 to 6 of 6

Math Help - Limit of an integral sequence

  1. #1
    Junior Member
    Joined
    Jan 2009
    From
    india
    Posts
    51

    Limit of an integral sequence

    how to solve
    If S= integral of n*x^(n-1)/(1+x) for n>=1 with limit 0 to 1
    then find limit of sequence {S} if n tends to infinity.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,373
    Thanks
    48
    Quote Originally Posted by Mathventure View Post
    how to solve
    If S= integral of n*x^(n-1)/(1+x) for n>=1 with limit 0 to 1
    then find limit of sequence {S} if n tends to infinity.
    Let I = \frac{\;\;nx^{n-1}}{1+x} and notice the following if we divide
    n = 1,\;\; I = \frac{1}{1+x}
    n = 2,\;\; I = 2\left( 1 - \frac{1}{1+x} \right)
    n = 3,\;\; I = 3\left( x - 1 + \frac{1}{1+x} \right)
    n = 4,\;\; I = 4\left( x^2 - x + 1 - \frac{1}{1+x} \right)
    n = 5,\;\; I = 5\left( x^3 - x^2 + x - 1 + \frac{1}{1+x} \right)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    From
    india
    Posts
    51
    Quote Originally Posted by danny arrigo View Post
    Let I = \frac{\;\;nx^{n-1}}{1+x} and notice the following if we divide
    n = 1,\;\; I = \frac{1}{1+x}
    n = 2,\;\; I = 2\left( 1 - \frac{1}{1+x} \right)
    n = 3,\;\; I = 3\left( x - 1 + \frac{1}{1+x} \right)
    n = 4,\;\; I = 4\left( x^2 - x + 1 - \frac{1}{1+x} \right)
    n = 5,\;\; I = 5\left( x^3 - x^2 + x - 1 + \frac{1}{1+x} \right)
    then what will be the exact answer and how....there r 4 options available
    0,1/2,1,infinity
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2008
    Posts
    96
    i got zero...if thats the answer, ill post the solution...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    From
    india
    Posts
    51
    Quote Originally Posted by adhyeta View Post
    i got zero...if thats the answer, ill post the solution...
    even i am nt sure of answer......whatever ur solution is, plz post it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2008
    Posts
    96

    Cool soln

    i had made a mistake. its 1/2

    here's my solution,



    Now, our aim is to get rid of the n in the first place...

    i used by parts for that.

    using by parts taking 1/(1+x) as the first function, we get;



    The first term of the sum has the limits 0->1 & we get rid of n from both terms by multiplying the n outside...

    So,



    Now, applying the limits of integration in the first terms gives us,



    Then,



    Now, observe the second term, here, x lies between 0 & 1. so x is basically a fraction & if we increase the power of a fraction, it decreses.

    If we take the power to infinity, it almost (tends to) 0.

    Hence, we have,



    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: October 26th 2010, 10:23 AM
  2. Sequence and limit
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: September 28th 2010, 07:55 PM
  3. Integral of a function as limit of a sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 24th 2010, 03:14 AM
  4. Limit of a Sequence
    Posted in the Differential Geometry Forum
    Replies: 30
    Last Post: March 6th 2010, 09:10 AM
  5. Limit of sequence
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 12th 2008, 03:16 PM

Search Tags


/mathhelpforum @mathhelpforum