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Limit of definite sum equals ln(2)

  1. Mar 10, 2013 #1
    1. The problem statement, all variables and given/known data
    As part of a problem I have to show that [tex]lim_{n\to\infty}\sum_{i=\frac{n}{2}}^{n}\frac{1}{i}=ln(2)[/tex]


    2. Relevant equations
    Taylor expansion of ln(2): [tex]\sum_{i=1}^{\infty}\frac{(-1)^{k+1}}{k}[/tex]


    3. The attempt at a solution
    ln(2) can be written as: [tex]ln(2) = \sum_{i=1}^{\frac{n}{2}}\frac{(-1)^{k+1}}{k} + \sum_{i=\frac{n}{2}+1}^{n}\frac{(-1)^{k+1}}{k} + \sum_{i=n+1}^{\infty}\frac{(-1)^{k+1}}{k}[/tex]

    Where the middle term looks alot like the sum i need. However I need some way to get rid of the alternating sign change, but i don't see how.
     
  2. jcsd
  3. Mar 10, 2013 #2

    Ray Vickson

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    Homework Helper

    You can use the exact result that
    [tex] \sum_{i=1}^n \frac{1}{i} = \Psi(n+1) + \gamma,[/tex]
    where ##\gamma## is Euler's constant and
    [tex] \Psi(x) = \frac{\Gamma^{\, \prime}(x)}{\Gamma(x)}[/tex]
    is the so-called "di-gamma" function. This function has numerous published properties, which should allow you to derive the answer you want.
     
  4. Mar 10, 2013 #3
    Thanx. But I think this approach is beyond the scope of my course. I was specifically told to look at the taylor expansion of ln(2). Is there a way to do it that way?
     
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