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Math Help - Telescoping sum of series

  1. #1
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    Telescoping sum of series

    Consider  a_n=\frac{1}{n^2+n}

    Find  S_3, S_6, and S_10

    \sum,(\frac{1}{n}+\frac{1}{(n+1)})
    n=\infty

    n=1  S_1 = (\frac{1}{1}+\frac{1}{1+1}) = (1+\frac{1}{2})

    n=2  S_2 = (\frac{1}{2}+\frac{1}{2+1}) = (\frac{1}{2}+\frac{1}{3})

    n=3  S_3 = (\frac{1}{3}+\frac{1}{3+1}) =  (\frac{1}{3}+\frac{1}{4})

    n=6  S_6 = (\frac{1}{6}+\frac{1}{6+1}) =   (\frac{1}{6}+\frac{1}{7})

    n=10  S_10 = (\frac{1}{10}+\frac{1}{10+1}) =   (\frac{1}{10}+\frac{1}{11})


    is that right?
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  2. #2
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    Quote Originally Posted by Anemori View Post
    Consider  a_n=\frac{1}{n^2+n}

    Find  S_3, S_6, and S_10

    \sum,(\frac{1}{n}+\frac{1}{(n+1)})
    n=\infty

    n=1  S_1 = (\frac{1}{1}+\frac{1}{1+1}) = (1+\frac{1}{2})

    n=2  S_2 = (\frac{1}{2}+\frac{1}{2+1}) = (\frac{1}{2}+\frac{1}{3})

    n=3  S_3 = (\frac{1}{3}+\frac{1}{3+1}) =  (\frac{1}{3}+\frac{1}{4})

    n=6  S_6 = (\frac{1}{6}+\frac{1}{6+1}) =   (\frac{1}{6}+\frac{1}{7})

    n=10  S_10 = (\frac{1}{10}+\frac{1}{10+1}) =   (\frac{1}{10}+\frac{1}{11})


    is that right?
    First, use Partial Fractions to simplify \frac{1}{n^2 + n} = \frac{1}{n(n + 1)}.


    If \frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}

    \frac{1}{n(n + 1)} = \frac{A(n + 1) + Bn}{n(n + 1)}.


    Then A(n + 1) + Bn = 1

    An + A + Bn = 0n + 1

    (A + B)n + A = 0n + 1.


    So A = 1 and A + B = 0.

    So B = -1.


    Therefore \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}.



    So for your series:

    \sum_{i = 1}^n{\frac{1}{i(i + 1)}} = \sum_{i = 1}^n{\left(\frac{1}{i} - \frac{1}{i + 1}\right)}.


    Notice that if we expand this out...

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

    Can you see that every term except for the first and last cancels?

    So the sum is 1 - \frac{1}{n + 1}.



    What do you think S_3, S_6 and S_{10} equal?


    P.S. A telescopic series is a series where everything except the first and last terms cancel.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    First, use Partial Fractions to simplify \frac{1}{n^2 + n} = \frac{1}{n(n + 1)}.


    If \frac{1}{n(n + 1)} = \frac{A}{n} + \frac{B}{n + 1}

    \frac{1}{n(n + 1)} = \frac{A(n + 1) + Bn}{n(n + 1)}.


    Then A(n + 1) + Bn = 1

    An + A + Bn = 0n + 1

    (A + B)n + A = 0n + 1.


    So A = 1 and A + B = 0.

    So B = -1.


    Therefore \frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}.



    So for your series:

    \sum_{i = 1}^n{\frac{1}{i(i + 1)}} = \sum_{i = 1}^n{\left(\frac{1}{i} - \frac{1}{i + 1}\right)}.


    Notice that if we expand this out...

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

    Can you see that every term except for the first and last cancels?

    So the sum is 1 - \frac{1}{n + 1}.



    What do you think S_3, S_6 and S_{10} equal?


    P.S. A telescopic series is a series where everything except the first and last terms cancel.
     S_3=\frac{3}{4}, S_6=\frac{6}{7} and S_10=\frac{10}{11}

    right?
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  4. #4
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    Quote Originally Posted by Anemori View Post
     S_3=\frac{3}{4}, S_6=\frac{6}{7} and S_10=\frac{10}{11}

    right?
    Correct.
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