1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes. I take it you've not done any integration yet.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted