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Finding explicit formula of this recursion formula

  1. May 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Write an explicit formula for the sequence determined by the following recursion formula.

    t[itex]_{1}[/itex]= 0; t[itex]_{n}[/itex] = t[itex]_{n-1}[/itex] + [itex]\frac{2}{n(n+1)}[/itex]




    3. The attempt at a solution

    t[itex]_{1}[/itex] = 0

    t[itex]_{2}[/itex] = t[itex]_{1}[/itex] + [itex]\frac{2}{2(2+1)}[/itex]
    t[itex]_{2}[/itex] = [itex]\frac{1}{3}[/itex]

    t[itex]_{3}[/itex] = t[itex]_{2}[/itex] + [itex]\frac{2}{3(3+1)}[/itex]
    t[itex]_{3}[/itex] = [itex]\frac{1}{3}[/itex] + [itex]\frac{2}{3(3+1)}[/itex]
    t[itex]_{3}[/itex] = [itex]\frac{4}{12}[/itex] + [itex]\frac{2}{12)}[/itex]
    t[itex]_{3}[/itex] = [itex]\frac{1}{2}[/itex]

    t[itex]_{4}[/itex] = t[itex]_{3}[/itex] + [itex]\frac{2}{4(4+1)}[/itex]
    t[itex]_{4}[/itex] = [itex]\frac{1}{2}[/itex] + [itex]\frac{2}{20}[/itex]
    t[itex]_{4}[/itex] = [itex]\frac{3}{5}[/itex]


    My sequence is 0, [itex]\frac{1}{3}[/itex], [itex]\frac{1}{2}[/itex], [itex]\frac{3}{5}[/itex] [itex]\cdots[/itex]

    How do I make an explicit formula if there is no common difference nor a common ratio?
     
  2. jcsd
  3. May 15, 2013 #2

    Curious3141

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    Do a partial fraction decomposition on ##\frac{2}{n(n+1)}##. Let that be ##\frac{A}{n} + \frac{B}{n+1}## (you determine A and B).

    Now ##t_n = t_{n-1} + \frac{A}{n} + \frac{B}{n+1}## and ##t_{n-1} = t_{n-2} + \frac{A}{n-1} + \frac{B}{n}##.

    Substitute the latter expression into the first and see what happens. Now continue successive substitution until you arrive at ##t_1##.
     
  4. May 15, 2013 #3

    haruspex

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    Here's another way. The form n(n+1)(n+2)... (n+r-1) in sums of series is strongly analogous to the form xr in integration. So for Ʃ1/(n(n+1)) consider ∫dx/x2. This gives you a guess for the sum of the series, which you can then refine by taking the difference of two consecutive terms and comparing it with the original.
     
  5. May 15, 2013 #4
    is that calculus? :$
     
  6. May 15, 2013 #5

    haruspex

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    Yes. I take it you've not done any integration yet.
     
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