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Given nth partial sum of a series, find a of n and sum

  1. Jul 4, 2016 #1
    1. The problem statement, all variables and given/known data
    If the nth partial sum of a series ##\sum_{n=1} ^\infty a_{n}## is
    ##S_{n} = \frac {n-1} {n+1}##
    Find ##a_{n}## and ##\sum_{n=1}^\infty a_n##

    2. Relevant equations
    ##S_{n} - S_{n-1}= a_{n}##
    ##\lim_{n \rightarrow +\infty} {S_{n}} = \sum_{n=1}^\infty a_n = S##

    3. The attempt at a solution
    So I used the first equation and I found ##a_{n} = \frac {2} {n(n+1)}##

    Work: ##S_{n} - S_{n-1}= a_{n}= \frac {n-1} {n+1} - \frac {n-2} {n} = \frac {(n-1)(n) - (n+1)(n-2)} {n(n+1)} = \frac {n^2 -n -n^2 - (-n -2)}{n(n+1)} = \frac {2} {n(n+1)}##

    The part I'm having a hard time understanding is the sum part.

    ##\lim_{n \rightarrow +\infty} {\frac {n-1} {n+1}} = 1##, however, ##\sum_{n=1}^\infty \frac {2} {n(n+1)} = 2(\sum_{n=1}^\infty \frac {1} {n(n+1)})##, and ##\sum_{n=1}^\infty \frac {1} {n(n+1)}## is a telescoping series that comes out as 1, so shouldn't the sum of the series equal 2? Why do these come out as 2 different answers even though they are suppose to be the same?

    The answer at the back of the book says ##a_{1} = 0##, and I have a hard time understanding this, as when n=1, ##a_{1} = 2/2 = 1##. If it's because you plug 1 into ##S_{n}## and you get 0, then that still leaves me confused, as ##S_{1} = a_{1}##, but it isn't, or is my understanding incorrect?
     
  2. jcsd
  3. Jul 4, 2016 #2
    The problem here is that ##a_{n} = \frac{2}{n(n+1)}## only for ##n\neq 1##. By using ##a_{n} = S_{n} - S_{n-1}##, you made the implicit assumption that ##S_{n-1}## exists, which for ##n = 1##, it clearly does not.
     
  4. Jul 4, 2016 #3
    Ahh, I see now; I had a feeling it was something I skipped over. Thanks for the help!
     
    Last edited: Jul 4, 2016
  5. Jul 4, 2016 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    The nth partial sum means ##S_n = \sum_{k=1}^n a_k##, so ##S_1 = a_1## and ##S_n = a_n + S_{n-1}## for ##n > 1##.
     
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