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Equality between sequences.

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that [tex]\sum_{k=1}^n \frac{(-1)^{k-1}}{k} \binom{n}{k} = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n-1} + \frac{1}{n} = \sum_{k=1}^n \frac{1}{k}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Writing out few of the summands:
    [tex]\frac{n!}{1\cdot 1!(n-1)!} - \frac{n!}{2\cdot 2!(n-2)!} + \frac{n!}{3\cdot 3!(n-3)!} - \frac{n!}{4\cdot 4!(n-4)!} +...\\
    n!(\frac{1}{1\cdot 1!(n-1)!}-\frac{1}{2\cdot 2!(n-2)!}+\frac{1}{3\cdot 3!(n-3)!}-\frac{1}{4\cdot 4!(n-4)!} + ...)[/tex]
    if this really adds up the way it's going to, I would somehow have to show that what is between the parenthesis adds up to [itex]\frac{1}{k\cdot n!}[/itex]then [itex]n!\cdot \frac{1}{k\cdot n!}[/itex] would be 1/k: what I am looking for.
    How should I proceed?
  2. jcsd
  3. Nov 25, 2014 #2


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    Staff: Mentor

    There is the related formula
    $$\sum_{k=0}^n \frac{(-1)^{k}}{k} \binom{n}{k} = \begin{cases}
    \frac{1}{n}, & \text{n odd}\\
    1-\frac{1}{n},& \text{n even} \end{cases}$$
    I wonder if you could prove both via induction (that's how I got that formula, using the result you want to show).
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