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Proof by induction

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]\forall n\in \mathbb{N}[/itex]
    [tex]\sum\limits_{k=1}^{n}(-1)^{k+1}{n\choose k}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}[/tex]

    2. Relevant equations
    -Induction
    -Summation
    -Binomial coefficient

    3. The attempt at a solution

    For [itex]n=1[/itex] equality is true.

    For [itex]n=m[/itex]

    [itex]m-{m\choose 2}\frac{1}{2}+...+(-1)^{m+1}\frac{1}{m}=1+\frac{1}{2}+...+\frac{1}{m}[/itex]

    For [itex]n=m+1[/itex]

    [itex]\left(\sum\limits_{k=1}^{m}(-1)^{k+1}{m\choose k}\frac{1}{k}\right)+(-1)^{m+2}\frac{1}{m+1}=1+\frac{1}{2}+...+\frac{1}{m+1}[/itex]

    If [itex]m[/itex] is even, equality is true, but not if [itex]m[/itex] is odd.
    Is this correct?

    Could someone explain this in detail?
     
  2. jcsd
  3. Oct 18, 2015 #2
    Hi gruba:

    You have made some mistakes.

    (1) The series being added does not include (-1)k+1. That series should be 1-1/2+1/3-1/4...

    (2) The series being added does not include (m over k) with each of the fractions.

    Hint: For (m over k) see: https://en.wikipedia.org/wiki/Combination

    I hope this helps.

    Regards,
    Buzz
     
  4. Oct 18, 2015 #3
    Are you sure it is a mistake?
    I have some solution which I don't inderstand:

    [itex]\sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1} x^k = 1-(1-x)^n\tag{1}[/itex]
    [itex]\sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{1}{k} = \int_{0}^{1}\frac{1-(1-x)^n}{x}\,dx= \int_{0}^{1}\frac{1-x^n}{1-x}\,dx\tag{2}[/itex]
    [itex]\sum_{k=1}^{n}\binom{n}{k}(-1)^{k+1}\frac{1}{k} = \int_{0}^{1}\left(1+x+\ldots+x^{n-1}\right)\,dx = 1+\frac{1}{2}+\ldots+\frac{1}{n}=H_n\tag{3}[/itex]
     
  5. Oct 18, 2015 #4
    Hi gruba:

    Your (1) is OK.
    The left hand equality in your (2) is OK. I don't see how you get the right hand equality in your (2).

    ADDED
    Ah. You substituted x -> (1-x). You forgot to also substitute dx -> -dx.

    Regards,
    Buzz
     
  6. Oct 18, 2015 #5
    This is my book's solution which I don't understand.
    Is there some approach which doesn't involve integration?
     
  7. Oct 18, 2015 #6
    Hi gruba:

    I edited a bit more to my last post after you responded. Please take another look at my previous post.

    Regards,
    Buzz
     
  8. Oct 18, 2015 #7
    Hi gruba:

    Sorry. My mistake. The missing minus sign is corrected by reversing the top and bottom limits of the integration.

    Regards,
    Buzz
     
  9. Oct 18, 2015 #8
    Hi gruba:

    My apologies for my earlier confusion.

    The derivation (1), (2) and (3) is a proof of the original equality. However it is not a proof by induction.

    Going back to your original post, after
    the (m over k) in the summation should be (m+1 over k).

    Regards,
    Buzz
     
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