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Sum as formula

  1. Oct 21, 2011 #1
    How can I show that [tex]\sum_{i=1}^n\;\frac1{i(i+1)}=\frac{n}{n+1}[/tex]

    I've already figured out i can write it as [tex]\sum_{i=1}^n\;\frac1{i}-\sum_{i=1}^n\;\frac1{i+1}[/tex]

    but as I'm a little drunk I can't figure out how to get from there to the formula.

    Sorry if I put this in the wrong sextion, but twas in my calculus book.
  2. jcsd
  3. Oct 21, 2011 #2


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    1) Don't drink and derive.

    2) What would happen, say, if you wrote the first sum as [itex]1 + \sum_{i=2}^n (1/i)[/itex] and then changed variables in that sum to something convenient?
  4. Oct 21, 2011 #3


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    It is called a telescoping series. The idea is only one term from each sum remain after all others cancel.
  5. Oct 22, 2011 #4


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    If I understand what you are asking, it is just
    [tex]\sum_{i= 1}^n\frac{1}{i}+ \sum_{i= 1}^n\frac{1}{i+ 1}= \sum_{i= 1}^n\left(\frac{1}{i}- \frac{1}{i+ 1}\right)[/tex]
    [tex]= \sum_{i= 1}^n \left(\frac{i+1}{i(i+1)}- \frac{i}{i(i+1)}\right)= \sum_{i= 1}^n\frac{1}{i(i+1)}[/tex]

    In other words, combine the two sums, get common denominators, and subtract the numerators.
  6. Oct 22, 2011 #5


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    Telescoping result. 1 - 1/(n+1) = n/(n+1)
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