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Finding limit of sum

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{n\to\infty}\frac{\ln 2 - \sum_{k=1}^{n}\frac{1}{k+n}}{\ln 2 - \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}}=? [/tex]


    2. Relevant equations
    [tex]H(2n)-H(n)= \sum_{k=1}^{n}\frac{1}{k+n}[/tex]

    3. The attempt at a solution
    I tried to use that [tex]\ln 2 = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{k}[/tex], but with no success. Can you help me, please?
     
  2. jcsd
  3. Feb 27, 2010 #2
    Re: Limit

    Any ideas?
     
  4. Feb 27, 2010 #3
    Re: Limit

    The sum term in the numerator approaches ln(2) as n goes to infinity since the sum can be manipulated into resembling a Riemann sum (via a regular n equal subdivisions partition of [0,1]). Also what exactly is H?
     
  5. Feb 27, 2010 #4
    Re: Limit

    yes, I already found that, but how does it help us?
    H - harmonic series: [tex]H(n)=\sum_{k=1}^{n}\frac{1}{k}[/tex]; It's not part of the problem, just thought it can be useful...
     
  6. Feb 27, 2010 #5

    Dick

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    Science Advisor
    Homework Helper

    Re: Limit

    You can write complicated expressions for the numerator and denominator in terms of the digamma function. Which means you can apply l'Hopital. What can do with that depends on how much you know about the digamma functions. I don't know much. Just playing with the numbers seems to show the absolute value of that ratio approaches a limit. But since the numerator is an alternating series approximation the sign keeps flipping.
     
    Last edited: Feb 27, 2010
  7. Feb 28, 2010 #6
    Re: Limit

    Sadly, I know nothing about the digamma function :(
     
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