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Prove sequence is bounded above

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Let a_n = 1 + 1/(1*2) + 1/(2*3) + ... + 1/(n*[n+1]). Prove {a_n} is bounded above.


    2. Relevant equations
    1/(2*3) = 1/2 - 1/3


    3. The attempt at a solution
    I accidentally left my notebook at school and I have no idea how to do this without my class notes. The book doesn't have any examples to help me out either. Please help...
     
  2. jcsd
  3. Jan 29, 2012 #2
    I'd prove the series converges and is monotonically increasing
     
  4. Jan 29, 2012 #3

    Dick

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    It's a telescoping series, isn't it?
     
  5. Jan 29, 2012 #4
    Use your "relevant equation" on the first five or six terms and see where Dick's term comes from...
     
  6. Jan 29, 2012 #5
    ok so this is what i have so far, i'm not sure if it "proves" it completely though:
    a1 = 1/1 - 1/2 = 1/(1*2)
    a2 = 1/2 - 1/3 = 1/(2*3)
    a3 = 1/3 - 1/4 = 1/(3*4)
    ....
    a(n-1) = 1/(n-1) - 1/n = 1/((n-1)*n)
    an = 1/n - 1/(n+1) = 1/(n*(n+1))

    Adding everything together
    (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/(n-1) - 1/n) + (1/n - 1/(n+1))
    1 - 1/(n+1) = 1/(1*2) + 1/(2*3) + ... + 1/(n*(n+1)) = an - 1
    Add 1 to both sides
    an = 2 - 1/(n+1) = (2n+1)/(n+1)
    lim(an) = 2
    Therefore the series is bounded above by 2

    does this complete my proof?
     
  7. Jan 29, 2012 #6

    Dick

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    I think that would do it, yes!
     
  8. Jan 30, 2012 #7
    Would it be overkill to prove [itex]a_{n}=2-\frac{1}{n+1}[/itex] by induction?

    Additionally my "quick and dirty" bounded-above argument would have been to observe that the terms are less than the reciprocals of squares (1/(2*3) < 1/(2*2) etc). Getting the least upper bound is a bonus, of course.
     
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