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Nth term of a series

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data
    Not really sure where tou go with this one.

    2. Relevant equations
    If the nth partial sum of a partial series is given by,

    Sn= [tex]\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}[/tex] = [tex]\sum_{k=1}^{n}[/tex][tex]\frac{1}{n+k}[/tex]

    a) write the associated series
    b) test for convergence
    c) if possible, determine its limit


    3. The attempt at a solution

    Here is what I have come up with:
    [tex]s_1[/tex]=1/2
    [tex]s_2[/tex]=1/2+1/3+1/4
    [tex]s_3[/tex]=1/2+1/3+1/4+1/5+1/6

    [tex]a_n = \left\{ \begin{array}{c} \frac{1}{2} \text{ for }n=1 \\ \frac{1}{2n-1}+\frac{1}{2n} \text{ for }n\geq 2 \end{array} \right.
    [/tex]

    I don't know what to do next. What do I do with the 1/2?

    I am pretty sure I can handle b and c, I just need help with a.

    Thanks in advance!
     
  2. jcsd
  3. Oct 27, 2008 #2

    Gib Z

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    Homework Helper

    You seem to have the general term. Really, you just now just say that associated series is sum of a_n, but if you want to put it "nicely", then putting in n=1 in the expression you have for n>2, its equal to 1.5, which is 1 too big. So you can use the same generating rule for all n, as long as you subtract 1 from the series.
     
  4. Oct 27, 2008 #3

    tiny-tim

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    "associated series"

    Hi Aimee79! :smile:

    The associated series just means the series {an} such that a1 + … + an = Sn.

    Hint: subtract something. :wink:
     
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