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Thread: Telescoping sum of series

  1. #1
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    Telescoping sum of series

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

    Find $\displaystyle S_3, S_6, and S_10 $

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

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

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

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

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

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


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

    Find $\displaystyle S_3, S_6, and S_10 $

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

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

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

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

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

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


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


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

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


    Then $\displaystyle A(n + 1) + Bn = 1$

    $\displaystyle An + A + Bn = 0n + 1$

    $\displaystyle (A + B)n + A = 0n + 1$.


    So $\displaystyle A = 1$ and $\displaystyle A + B = 0$.

    So $\displaystyle B = -1$.


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



    So for your series:

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


    Notice that if we expand this out...

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

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

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



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


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


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

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


    Then $\displaystyle A(n + 1) + Bn = 1$

    $\displaystyle An + A + Bn = 0n + 1$

    $\displaystyle (A + B)n + A = 0n + 1$.


    So $\displaystyle A = 1$ and $\displaystyle A + B = 0$.

    So $\displaystyle B = -1$.


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



    So for your series:

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


    Notice that if we expand this out...

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

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

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



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


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

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

    right?
    Correct.
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