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How to evaluate what a series converges to?

  1. Aug 7, 2010 #1
    I was asked to evaluate the summation of [tex]\frac{1}{n(n+1)}[/tex] from n=1 to infinity

    I used partial fractions to obtain [tex]\frac{1}{n}[/tex] - [tex]\frac{1}{n+1}[/tex]

    From here I don't understand how to evaluate. In my solutions manual, they plugged in values from 1 to infinity showing (1 - 1/2+ (1/2 - 1/3)...etc and created a new series called Sn = 1 - [tex]\frac{1}{n+1}[/tex] then took the limit of that to infinity to get the answer 1.

    How would I know what Sn should be?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Aug 7, 2010 #2


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    So you want to evaluate:

    [tex]\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=1}^\infty \frac{1}{n+1}[/tex]

    One way to evaluate it is to consider what happens if you change variables in the second sum to m = n + 1. If you make this change of variables, do you see how you can deduce the solution?
  4. Aug 7, 2010 #3


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    Your series partial sum is (1-1/2)+(1/2-1/3)+...+(1/n-1/(n+1)). Regroup that like this 1+(-1/2+1/2)+(-1/3+1/3)+...+(-1/n+1/n)-1/(n+1). Lots of stuff cancels.
  5. Aug 7, 2010 #4


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    Quick check -- you realize they are (presumably) using Sn for the n-th partial sum? And remember that an infinite sum is the limit of the partial sums?
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