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Prove that ?

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that 1+1+1/2!+1/3!+....+1/n! < 3


    2. Relevant equations
    None


    3. The attempt at a solution
    Any suggestion how can I start?
    I dont want the solution that 1+1+1/2!+1/3!+...+1/n!=(1+1/n)^n and lim (1+1/n)^n = e<3 since the 2nd part is to prove (1+1/n)^n <3 . I don't want a solution. I just want to know how to start with it.
     
  2. jcsd
  3. Feb 18, 2012 #2

    tiny-tim

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    welcome to pf!

    hi topengonzo! welcome to pf! :smile:

    hint: can you see a way to prove that 1/2!+1/3!+....+1/n! < 1 ? :wink:
     
  4. Feb 18, 2012 #3
    hi tiny-tim

    this is the first thing i thought of and i actually worked without the first 2 terms. Prove by induction on this problem is impossible i think. So i dont have a clue how to solve it. :S
     
  5. Feb 18, 2012 #4

    tiny-tim

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    1/2+1/2.3+1/2.3.4+....+ < ? :wink:
     
  6. Feb 18, 2012 #5
    hint: does the original series look familiar to you at all? Have you compared it to any series you already know?
    (If you haven't learned taylor expansion ignore what I said here)
     
  7. Feb 18, 2012 #6
    i dont get what u mean. I am thinking of proving that sum of next terms is less than equal to current term that is:
    (1/3!) + (1/4!)+(1/5!)+....(1/infinity!)<(1/2!)
    (1/4!) + (1/5!)+(1/6!)+....(1/infinity!)<(1/3!)
    .... (1/(n+1)!) + (1/(n+2)!)+.... (1/infinity!)<(1/n!)

    Should I reach a solution with this method or can I solve it this way?
     
  8. Feb 18, 2012 #7
    I think its taylor expansion will give me e<3 and solved. Is there another way to solve it?
     
  9. Feb 18, 2012 #8
    deleted
     
    Last edited: Feb 18, 2012
  10. Feb 18, 2012 #9

    tiny-tim

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    sunjin09, please don't give the full answer :redface:
     
  11. Feb 18, 2012 #10
    THANK YOU VERY VERY MUCH!

    So I prove 1/k! < 1/ 2^(k-1) for k>=3 by induction and then i can say my series < 1+ 1 + 1/2 + 1/4 + 1/8 + ... which is geometric series with r=1/2 and first term 1
    implies 1+ 1/(1-1/2) = 3
     
  12. Feb 18, 2012 #11
    That is the same as [itex]\dfrac{1}{2} \ + \ \dfrac{1}{2}\cdot 3 \ + \ \dfrac{1}{2}\cdot3\cdot4 \ \ + \ ... \ + \ < \ ?[/itex]



    (You're using the dots for multiplication. Written out horizontally,
    grouping symbols are needed.)



    Instead, it could be correctly shown horizontally as:


    1/2 + 1/(2*3) + 1/(2*3*4) + ... + < ?


    or as



    [itex]1/2 \ + \ 1/(2\cdot 3) \ + \ 1/(2\cdot3\cdot4) \ + \ ... \ + \ < \ ?[/itex]
     
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