Results 1 to 7 of 7

Math Help - Summation of 1/i^2

  1. #1
    rxt5972
    Guest

    Summation of 1/i^2

    Hi guys,
    The summation of i^2= n*(n+1)*(2n+1)/6
    What is the summation od 1/i^2 ?
    Thanks for your help.
    rxt5972
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by rxt5972 View Post
    Hi guys,
    The summation of i^2= n*(n+1)*(2n+1)/6
    What is the summation od 1/i^2 ?
    Thanks for your help.
    rxt5972
    I do not think there is one.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    I saw your other double post. I agree with the response there.
    You have been given an honest and correct answer,

    However, I will give you a different answer.
    The infinite series (sum) is: \sum\limits_{k = 1}^\infty  {\frac{1}{{k^2 }}}  = \frac{{\pi ^2 }}{6}.
    Therefore, for large values of N we can use  \frac{{\pi ^2 }}{6} as an approximation of \sum\limits_{k = 1}^N  {\frac{1}{{k^2 }}}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Plato View Post
    I saw your other double post. I agree with the response there.
    You have been given an honest and correct answer,

    However, I will give you a different answer.
    The infinite series (sum) is: \sum\limits_{k = 1}^\infty  {\frac{1}{{k^2 }}}  = \frac{{\pi ^2 }}{6}.
    Therefore, for large values of N we can use  \frac{{\pi ^2 }}{6} as an approximation of \sum\limits_{k = 1}^N  {\frac{1}{{k^2 }}}.
    Let me tell you why I do not think so.

    Consider a similar example,
    \sum_{k=1}^n \frac{1}{k}

    And consider a Riemann integral over some positive closed interval on the function f(x)=1/x. If we try to calculate this sum directly through the definition of the Riemann integral we will probably not be able to because the true values is the logarithmic function which is transendental.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    Let me tell you why I do not think so.
    Consider a similar example,
    \sum_{k=1}^n \frac{1}{k}
    And consider a Riemann integral over some positive closed interval on the function f(x)=1/x. If we try to calculate this sum directly through the definition of the Riemann integral we will probably not be able to because the true values is the logarithmic function which is transendental.
    What in the world are you on about?
    The harmonic series, \sum_{k=1}^n \frac{1}{k}, is a very slowly divergent series.
    The series \sum_{k=1}^n \frac{1}{k^2} converges rather rapidly.
    If fact for N=100 the estimate is correct in the first two decimal places.
    For N=1000 the estimate is correct in the first four decimal places.

    What is you point?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Plato View Post

    What is you point?
    I was working on a finite interval. So if you use a Riemann sum you need a way to add a finite harmonic series. Which will lead to a transendental function (the natural logarithm).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    I was working on a finite interval. So if you use a Riemann sum you need a way to add a finite harmonic series. Which will lead to a transendental function (the natural logarithm).
    If N=1000000 the harmonic sum is less than 15.
    But \sum_{k=1}^n \frac{1}{k^2} is correct in first six decimal places.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to do a Summation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 14th 2011, 12:08 PM
  2. Summation
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: May 7th 2011, 07:23 AM
  3. summation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 2nd 2009, 09:17 AM
  4. Summation Help
    Posted in the Algebra Forum
    Replies: 9
    Last Post: January 31st 2009, 08:47 PM
  5. summation
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 8th 2008, 05:19 PM

Search Tags


/mathhelpforum @mathhelpforum