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How to find Sn from Un? (sum of series)

  1. Sep 21, 2015 #1
    How to derive Sn (sum formula of series) from Un (nth-term of series) ?

    In my textbook (in the Mathematical Induction chapter), it's shown that
    ##\frac{1}{1\times 3}+\frac{1}{3\times5}+\frac{1}{5\times7}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}##

    I know how to proof it by mathematical induction,

    But, I want to know how to derive the Sn (sum formula of series) from Un..
    I mean, how to get Sn = ##\frac{n}{2n+1}##
    From Un = ##\frac{1}{(2n-1)(2n+1)}## ??
     
  2. jcsd
  3. Sep 21, 2015 #2

    Svein

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    Just a small hint: [itex]\frac{1}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1} [/itex]. Now insert n=1 and n=2 and add...
     
  4. Sep 21, 2015 #3
    There is a factor of 2 missing, isn't it?
     
  5. Sep 21, 2015 #4

    Svein

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    Yes, of course. I could pretend that it really was to test you, but I plain forgot. The long and short of it is that the parts cancel out and you are left with the first and last element.
     
  6. Sep 22, 2015 #5
    But, ##\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{2}{(2n-1)(2n+1)}## not ##\frac{1}{(2n-1)(2n+1)}##

    However, I can write it ##\frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\times(\frac{1}{2n-1}-\frac{1}{2n+1})##

    Then, what should I do to determine the Sn formula??

    For n = 1, then U1= 1/3
    For n = 2, then U2= 1/15
    For n = 3, then U3 = 1/35

    However, the difference between term 1 and term 2 is not the same as the difference between term 2 and term 3
    Also, the ratio is different
    How to determine Sn??
     
  7. Sep 22, 2015 #6

    Svein

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    As I said, write it out for n=1, 2, 3. This gives: [itex]S_{3}=\frac{1}{2}(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}) [/itex]. Do you see a pattern?
     
  8. Sep 22, 2015 #7
    Yupp
    It seems like Sn = ## \frac{1}{2}(1-\frac{1}{2n+1})##
    Simplified it Sn = ##\frac{n}{2n+1}##

    Thanks for your help!
     
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