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Math Help - Sequence and series

  1. #1
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    Sequence and series

    Q- Find the sum of the series -

    1+1/3+1/6+1/10+1/15+...........

    what should be our approach to solve these kind of questions ?

    Thanks,
    Ashish
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  2. #2
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    Hi

    S = \frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3}+\cdots+\frac{1}{1+2+\cdots+n}+\cdo  ts

    I presume that you know that 1+2+\cdots+n = \frac{n(n+1)}{2}

    Therefore

    S = \sum_{k=1}^{+\infty}\frac{2}{k(k+1)}

    Spoiler:

    S = 2\:\sum_{k=1}^{+\infty} \left(\frac{1}{k}-\frac{1}{k+1}\right) = 2
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  3. #3
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by a69356 View Post
    Q- Find the sum of the series -

    1+1/3+1/6+1/10+1/15+...........

    what should be our approach to solve these kind of questions ?

    Thanks,
    Ashish
    Hi again Ashish,


    I think you must have guessed that

    the general term of this series is \frac{2}{n(n+1)}


    You need to find

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


    Now all we are doing here is write every term

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

    as

    \frac{2((n+1)-(n))}{n(n+1)}

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


    this will make many terms cancel like the previous question

    -----------------------------------

    Final answer is

    Spoiler:
    2n/(n+1)


    use similar approach
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  4. #4
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    Sum of series

    Hello Ashish
    Quote Originally Posted by a69356 View Post
    Q- Find the sum of the series -

    1+1/3+1/6+1/10+1/15+...........

    what should be our approach to solve these kind of questions ?

    Thanks,
    Ashish
    First, you need to determine a formula for the n^{th} term of the series.

    The denominators are 1, 3, 6, 10, 15, ... whose consecutive differences are 2, 3, 4, 5, ...

    These are simply the sums of the series 1, 1+2, 1+2+3, 1+2+3+4, ... So the n^{th} term is the sum of the series 1+2+...+n = \tfrac12n(n+1)

    So the original series is:

    1 + \frac{1}{3}+ \frac{1}{6} + ... + \frac{1}{\tfrac12n(n+1)} + ...

    Using \Sigma notation, then, we have to find S=\sum_{n=1}^\infty\frac{2}{n(n+1)}

    This you do using Partial Fractions: \frac{2}{n(n+1)}=\frac{2}{n}-\frac{2}{n+1}

    \Rightarrow S = \sum_{n=1}^\infty\left(\frac{2}{n}-\frac{2}{n+1}\right)

    = \frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+ ...

    = 2

    Grandad
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  5. #5
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    Thanks a lot to all of you for quick reply....
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