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Thread: 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

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

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

    Therefore

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

    Spoiler:

    $\displaystyle S = 2\:\sum_{k=1}^{+\infty} \left(\frac{1}{k}-\frac{1}{k+1}\right) = 2$
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  3. #3
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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 $\displaystyle \frac{2}{n(n+1)} $


    You need to find

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


    Now all we are doing here is write every term

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

    as

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

    $\displaystyle = \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 $\displaystyle n^{th}$ term of the series.

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

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

    So the original series is:

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

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

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

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

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

    $\displaystyle = 2$

    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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