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Can I prove this?

  1. Feb 18, 2012 #1
    summation from k to n of 1/((k-1)!(k+1)) < 1/k! using induction

    How can i write it to look math
     
  2. jcsd
  3. Feb 18, 2012 #2

    mathman

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    You have 2 integer variables, k and n. What is the induction variable?
     
  4. Feb 18, 2012 #3
    n is induction variable and base term is n=k
     
  5. Feb 18, 2012 #4

    Fredrik

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    When you said "summation from k to n", I assume you meant "summation from 1 to n". For all induction problems, keep in mind that the goal is to prove infinitely many statements with a finite amount of work. If we for each positive integer n denote the nth statement by P(n), the goal is to prove P(1), P(2), P(3) and so on, by only proving the following statements:

    P(1)
    For all positive integers j, if P(j), then P(j+1).
     
  6. Feb 18, 2012 #5
    I assume the summation is from j to n, and that it ends with < 1/j! . It is not neccessary to state the induction variable, but it is the lower limit j. It is btw much easier to just compute the sum without any induction nonsense.
     
  7. Feb 18, 2012 #6

    D H

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    This is how I would interpret what you wrote:
    [tex]\sum_{r=k}^n \frac 1 {(k-1)!(k+1)} < \frac 1 {k!}[/tex]
    This is "the sum from k to n of 1/((k-1)!(k+1))" -- and it is false.


    This is what I think you meant:
    [tex]\sum_{r=k}^n \frac 1 {(r-1)!(r+1)} < \frac 1 {k!}[/tex]
    This is true for all finite integers k>0, n≥k. You don't need recursion to prove this.
     
  8. Feb 18, 2012 #7
    Yes the second one you wrote is what i exactly mean. How do I prove it? Also I think if i set n -> infinity (find lim at infinity), I would get = instead of < . Am I correct?
     
  9. Feb 18, 2012 #8

    D H

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    This is the homework section of PhysicsForums. You need to show some work.

    I will give a hint: Induction is not the way to go here. Simply find the value of the series.
     
  10. Feb 18, 2012 #9
    This is not the question of my homework. I will write it here down to show you that I am doing an effort and what I am asking will lead me to the answer.

    Question:
    Prove every positive rational number x can be expressed in ONE way in the form
    x= a1 + a2/2! + a3/3! + .... + ak/k!
    where a1,a2,...,ak are integers and 0<=a1,0<=a2<2,....,0<=ak<k
    My solution:
    To prove that there is only 1 solution, I am trying to prove
    min(ak/k!)>max( (a(k+1))/(k+1)!+(a(k+2))/(k+2)!+....+ (a(k+n))/(k+n)! )
    To get min set ak=1
    and to get max set ak=k

    and I get to the original question I am asking
     
  11. Feb 18, 2012 #10

    D H

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    I'll give you a start. We agree that you are trying to prove
    [tex]\sum_{r=k}^n \frac 1 {(r-1)!(r+1)} < \frac 1 {k!}[/tex]
    Multiplying each term of the sum by one won't change anything, but write one as r/r.
     
  12. Feb 18, 2012 #11
    I think that your previous answer using series will solve it.

    I am thinking of putting

    [tex]\sum_{r=k}^n \frac r {(r+1)!} < \frac 1 {k!}[/tex]



    and I use the poisson distribution series to prove = 1/k! and thus the sum of any finite terms of the series is < 1/k!

    Am I correct?
     
    Last edited: Feb 18, 2012
  13. Feb 18, 2012 #12

    D H

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    You're thinking too hard. Just rewrite the series so the sum goes from k+1 to n+1.
     
  14. Feb 18, 2012 #13
    [tex]\sum_{r=k+1}^{n+1} \frac {r-1} {(r)!} < \frac 1 {k!}[/tex]
    Is this correct?
    [tex]\sum_{r=0}^{inf} \frac {r-1} {(r)!}[/tex] = e-e=0

    And then I take out term from 0 to k?
     
  15. Feb 18, 2012 #14

    D H

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    Yes!

    No!

    You are thinking too hard! Look at the numerator. Rewrite that finite sum as the difference between two sums. Those two sums will have n-1 common terms. Get rid of them.
     
  16. Feb 18, 2012 #15
    [tex]\sum_{r=k+1}^{n+1} \frac {1} {(r-1)!} - \sum_{r=k+1}^{n+1} \frac {1} {(r)!} [/tex]

    YOUR RIGHT! it becomes simply
    [tex]\frac {1} {(k)!} - \frac {1} {(n+1)!} [/tex]

    Just 1 final question?

    I take lim as n tends to infinity of my answer so I get that the infinite sum of all terms after ak/k! is equal to ak/k! which means there is no overlapping and one and only one way to represent x.
    Also the sum of fractions span from 0 to infinity which means EVERY positive rational number x can be represented by the sum of fractions
     
  17. Feb 18, 2012 #16
    Is my solution correct and is there another way to solve it?
     
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