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Series problem: given s(sub n), find a(sub n) and sum of a(sub n)

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    This is a problem in my book, and the answer is in the back. Unfortunately I can't solve it.

    If the nth partial sum of a series [tex]\sum a_{n}[/tex] n=1 to infinity is

    [tex]s_{n} = (n - 1) / (n + 1)[/tex]

    find [tex]a_{n}[/tex] and [tex]\Sigma a_{n}[/tex] n=1 to infinity

    2. Relevant equations
    [tex]a_{n} = s_{n} - s_{n-1}[/tex]


    3. The attempt at a solution
    I used the equation above to get [tex]a_{n} = 2/(n(n+1))[/tex]

    To get the sum, I wrote out the sequence after I figured out:
    2/(n(n+1)) = 2/n - 2/(n+1)
    so
    = lim n->infinity (2/1 - 2/2) + (2/2 - 2/3) + (2/3 - 2/4) + ... + (2/n - 2/(n+1))
    = lim n->infty 2/1 - 2(n+1)
    = 2

    But the answer for the sum is 1.
     
  2. jcsd
  3. Nov 15, 2009 #2
    Not all a_n follow that formula -- check where it's not applicable.
     
  4. Nov 15, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The sum of an infinite series is, by definition, the limit of the sequence of partial sums. Here it is [itex]\lim_{n\to\infty} (n-1)/(n+1)[/itex].

    To find the individual terms, [itex]a_1[/itex] is the first "partial sum": [itex]a_1= (1-1)/(1+1)= 0[/itex]. [itex]a_2[/itex] is then given by [itex]a_1+ a_2= (2-1)/(2+1)[/itex] so [itex]0+ a_2= a_2= 1/3[/itex]. [itex]a_3[/itex] is given by [itex]a_1+ a_2+ a_3= 0+ 1/3+ a_3[/itex][itex]= (3-1)(3+1)= 2/4= 1/2[/itex] so [itex]a_3= 1/2- 1/3= 1/6[/itex], etc.
     
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