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Math Help - sum of 1st nth terms

  1. #1
    Member SengNee's Avatar
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    sum of 1st nth terms

    If the rth term of the series
    \frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+... is \frac{1}{1+2+3+...+r}
    show that the sum of the first n terms of the series is \frac{2n}{n+1}.
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  2. #2
    Member SengNee's Avatar
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    Quote Originally Posted by janvdl View Post
    You have the first and nth terms.

    S_{n} = n \left( \frac{a + l}{2} \right)

    Where a is the first term and l the last(or the nth term in this case).
    This formula is foe A.P.
    The series is A.P.?
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  3. #3
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    Quote Originally Posted by SengNee View Post
    The series is A.P.?
    Of course it is not A.P.
    Because a_r  = \frac{1}{{1 + 2 +  \cdots  + r}} = \left( {\sum\limits_{k = 1}^r k } \right)^{ - 1}  = \frac{2}{{r\left( {r + 1} \right)}}
    we get \sum\limits_{k = 1}^n {a_k }  = 2\sum\limits_{k = 1}^r {\left( {\frac{1}{k} - \frac{1}{{k + 1}}} \right) = \frac{{2n}}{{n + 1}}} <br />
.
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    \frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)}

    This means you want to sum, \sum_{k=1}^n \frac{2}{k(k+1)} = 2 \sum_{k=1}^n \frac{1}{k(k+1)}.

    Now, \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.

    Thus, 2\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = 2\left( 1 - \frac{1}{n+1} \right) = \frac{2n}{n+1}.
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  5. #5
    Member SengNee's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    \frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)}

    This means you want to sum, \sum_{k=1}^n \frac{2}{k(k+1)} = 2 \sum_{k=1}^n \frac{1}{k(k+1)}.

    Now, \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}.

    Thus, 2\sum_{k=1}^n \frac{1}{k} - \frac{1}{k+1} = 2\left( 1 - \frac{1}{n+1} \right) = \frac{2n}{n+1}.
    Why \frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)} ?
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  6. #6
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    Quote Originally Posted by SengNee View Post
    Why \frac{1}{1+2+\cdots + k} = \frac{2}{k(k+1)} ?
    Because 1+\cdots + k = \frac{k(k+1)}{2} --> arithmetic series.
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  7. #7
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    Hello, SengNee!

    Given the series: . S \;=\;\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+ \hdots +\frac{1}{1+2+3+\hdots +n} + \hdots

    show that the sum of the first n terms of the series is: . S_n \;=\;\frac{2n}{n+1}
    This is neither an Arithmetic Series nor a Geometric Series.
    . . We must find another approach . . .

    The denominators are "triangular numbers" ... of the form: . \frac{n(n+1)}{2}

    We have: . S_n \;=\;\frac{2}{1\cdot2} + \frac{2}{2\cdot3} + \frac{2}{3\cdot4} + \frac{2}{4\cdot5} + \hdots + \frac{2}{n(n+1)}

    . . . . . . . . . . =\;2\left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \hdots + \frac{1}{n(n+1)}\right]

    . . . . . . . . . . = \;2\left[\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \hdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right]


    All the "middle" terms cancel out and we are left with:

    . . . . . S_n \;\;=\;\;2\left[1 - \frac{1}{n+1}\right] \;\;=\;\;\frac{2n}{n+1}

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  8. #8
    Member SengNee's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, SengNee!

    This is neither an Arithmetic Series nor a Geometric Series.
    . . We must find another approach . . .

    The denominators are "triangular numbers" ... of the form: . \frac{n(n+1)}{2}

    We have: . S_n \;=\;\frac{2}{1\cdot2} + \frac{2}{2\cdot3} + \frac{2}{3\cdot4} + \frac{2}{4\cdot5} + \hdots + \frac{2}{n(n+1)}

    . . . . . . . . . . =\;2\left[\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \hdots + \frac{1}{n(n+1)}\right]

    . . . . . . . . . . = \;2\left[\left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \hdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)\right]


    All the "middle" terms cancel out and we are left with:

    . . . . . S_n \;\;=\;\;2\left[1 - \frac{1}{n+1}\right] \;\;=\;\;\frac{2n}{n+1}




    For triangular number,
    T_n=\frac{n}{2}(n+1)
    T_n +T_{n-1}=n^2

    Therefore, is it
    S_n= T_1+ T_2+ T_3+ \cdots +T_{n-1}+ T_n
    S_n=\frac{n}{2}(n^2)
    ?
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  9. #9
    Senior Member JaneBennet's Avatar
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    S_1=\frac{2(1)}{1+1}

    If S_k=\frac{2k}{k+1} then

    S_{k+1}=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{  2k(k+2)+2}{(k+1)(k+2)}=\frac{2(k^2+2k+1)}{(k+1)(k+  2)}=\frac{2(k+1)}{k+2}
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  10. #10
    Member SengNee's Avatar
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    Quote Originally Posted by JaneBennet View Post
    S_1=\frac{2(1)}{1+1}

    If S_k=\frac{2k}{k+1} then

    S_{k+1}=\frac{2k}{k+1}+\frac{2}{(k+1)(k+2)}=\frac{  2k(k+2)+2}{(k+1)(k+2)}=\frac{2(k^2+2k+1)}{(k+1)(k+  2)}=\frac{2(k+1)}{k+2}
    I mean triangular number, 1,3,6,10,15,..., not the series 1st I had posted.
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