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Sequence problem

  1. Dec 3, 2007 #1

    tony873004

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    1. The problem statement, all variables and given/known data
    Find the Sum of [tex]\sum\limits_{n = 2}^\infty {\frac{1}{{n\left( {n - 1} \right)}}} [/tex]



    2. Relevant equations
    [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}} = 1} [/tex]



    3. The attempt at a solution
    [tex]\sum\limits_{n = 2}^\infty {\frac{1}{{n\left( {n + 1} \right)}} = \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)}} - a_1 = 1} - a_1 = 1 - \frac{1}{{1\left( {1 + 1} \right)}} = 1 - \frac{1}{2} = \frac{1}{2}}
    [/tex]

    But the book says the answer is 1. I can see it being 1 if n=1 in the sumation symbol, but it is n=2, so don't I have to subtract the value of a1 from the sum computed from the formula?
     
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  3. Dec 3, 2007 #2

    morphism

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    Why are you finding \sum 1/n(n+1) when the question is asking for \sum 1/n(n-1)? (Note the difference in signs.)
     
  4. Dec 3, 2007 #3
    You may want to look up partial fraction decomposition. I will explain how you can compute the formula that is given to you (#2). The sum you're interested in can be computed in a similar fashion.

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

    There is a method for obtaining the right hand side given any rational expression -- you should look this up so that you understand how to apply this idea if you are given more complicated rational expressions as your n-th term in the series.

    Now, since

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

    try writing out a few terms and you should see a pattern. Try to apply this approach to your summation. (Note, as morphism already pointed out, that your summation has subtraction in the denominator, and not addition.)

    This type of summation/series is called a telescoping series.
     
    Last edited: Dec 3, 2007
  5. Dec 3, 2007 #4

    tony873004

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    Thanks, I mixed up a + and - sign.
     
  6. Dec 3, 2007 #5

    tony873004

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    [tex]\sum\limits_{n = 2}^\infty {\frac{1}{{n\left( {n - 1} \right)}} = \sum\limits_{n = 1}^\infty {\frac{1}{{n - 1}} - \frac{1}{n} = \left( {\frac{1}{1}} \right)} } + \left( {} \right) + \left( {} \right)...\left( { - \frac{1}{n}} \right) = \mathop {\lim }\limits_{n \to \infty } \frac{1}{1} - \frac{1}{n} = 1 - 0 = 1[/tex]

    Everything but the first and last terms cancel, so now I get 1. But I still don't understand decomposition.

    How should I have known to do this:
    [tex]\frac{1}{{n\left( {n - 1} \right)}} = \frac{1}{{n - 1}} - \frac{1}{n}[/tex]
     
  7. Dec 3, 2007 #6
    You have a minor mistake:

    [tex]\sum\limits_{n = 2}^\infty {\frac{1}{{n\left( {n - 1} \right)}} = \sum_{ n=\mathbf{2}}^\infty \left({\frac{1}{{n - 1}} - \frac{1}{n} \right)[/tex]

    Regarding the decomposition, look up partial fractions. It's essentially the reverse of combining two fractions. Judging by your response, perhaps partial fractions have not been covered. Fortunately, there is another method which DOES use the formula in (#2) directly.

    Write out a few terms of the sum

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

    and compare with the terms from

    [tex]\sum_{n=2}^\infty \frac{1}{n(n-1)} = \sum_{n=2}^\infty \frac{1}{(n-1)n}[/tex]

    (Use the form on the right hand side. Just write out the terms explicitly -- don't simplify)

    NOTE THE INDICES; THEY ARE DIFFERENT.
     
  8. Dec 3, 2007 #7

    tony873004

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    That mistake was a typo. Thanks for catching it. I think we covered partial fractions in the week I was sick. I knew it'd come back to bite me. Thanks for your help! And thanks for the alternate method.
     
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