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Math Help - Writing a sequence in closed form

  1. #1
    Newbie madmartigano's Avatar
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    Writing a sequence in closed form

    Consider the sequence \left\{ {{a_n}} \right\}_{n = 1}^{ +  \infty }, where

    {a_n} = \frac{1^2}{{{n^3}}} + \frac{2^2}{{{n^3}}} +  \cdots  + \frac{n^2}{{{n^3}}}

    I've written out enough terms to guess that this sequence converges at 1/3, but I'm supposed to confirm my conjecture by expressing {{a_n}}<br />
in closed form and calculating the limit.

    I understand the method for finding the sum of

    {a_n} = 1 + 2 +  \cdots  + n, which is \frac{{n(n + 1)}}{2} in closed form,

    but I'm having trouble figuring out an equivalent equation for

    {a_n} = {1^2} + {2^2} +  \cdots  + {n^2}

    Any help would be very much appreciated.
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    \sum\limits_{k = 1}^n {k^2 }  = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}<br />
{6}
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  3. #3
    Newbie madmartigano's Avatar
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    Thank you, Plato, but how did you arrive at that conclusion?
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    Quote Originally Posted by madmartigano View Post
    Thank you, Plato, but how did you arrive at that conclusion?
    That is just a widely known fact.
    One finds it in any calculus textbook.
    Or prove it by induction.
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  5. #5
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    Quote Originally Posted by madmartigano View Post
    Thank you, Plato, but how did you arrive at that conclusion?
    Here's a way. Consider

    \sum_{i=1}^n (i+1)^3 - i^3

    If you write out the terms

    \sum_{i=1}^n (i+1)^3 - i^3 = \left( 2^3-1^3\right) + \left( 3^3-2^3\right) + \left( 4^3-3^3\right) + \cdots + \left( (n+1)^3-n^3\right)

    and you see all the terms cancel except two so

    \sum_{i=1}^n (i+1)^3 - i^3 = (n+1)^3-1 = n^3 + 3n^2 + 3n

    Now expand the sum

    \sum_{i=1}^n (i+1)^3 - i^3 = \sum_{i=1}^n i^3 + 3i^2 + 3i + 1 - i^3 = 3\sum_{i=1}^n i^2 + \underbrace{3\sum_{i=1}^n i + \sum_{i=1}^n 1}_{\text{we know these}} = 3\sum_{i=1}^n i^2 + \frac{3}{2} n (n+1) + n

    Now equate each part

     3\sum_{i=1}^n i^2 + \frac{3}{2} n (n+1) + n = n^3 + 3n^2 + 3n

    Solving for the sum gives, nothing up my sleeve, presto

    \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
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