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[itex]\lim_{n \to \infty} \sum_{1}^{n} 1/(n+i)=log(2)[/itex]

  1. Aug 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove that [itex]\lim_{n \to \infty} \displaystyle \sum_{i=1}^{n} \frac{1}{n+i} = log(2)[/itex].

    2. Relevant equations

    The Digamma Function [itex]\Psi(x)[/itex] can be written as:

    [itex]\Psi (x) = \frac{d}{dx} \log(\Gamma(x))[/itex]

    Or, equivalently, as

    [itex]\Psi(x)=\displaystyle \sum_{l=1}^{k} \frac{1}{x-l}+\Psi(x-k)[/itex]

    3. The attempt at a solution
    I tried to put in Wolfram in order to get some hint about this problem and it showed that:

    [itex]\displaystyle \sum_{i=1}^{n} \frac{1}{n+i} = \Psi(n+(i+1))-\Psi(1+i)[/itex]

    Where [itex]\Psi (x) = \frac{d}{dx} \log(\Gamma(x))[/itex] is the digamma function as described in Relevant equations.
     
  2. jcsd
  3. Aug 6, 2017 #2

    pasmith

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    [itex]i[/itex] occurs as a summation index on the left. It therefore cannot appear as part of the result of doing the sum. So something is not correct here.

    The digamma function is defined in terms of the gamma function, which is the function which satisfies the recurrence relation [tex]
    \Gamma(z + 1) = z \Gamma (z)[/tex] subject to [itex]\Gamma (1) = 1[/itex].
     
    Last edited: Aug 6, 2017
  4. Aug 6, 2017 #3

    Ray Vickson

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    No: Wolfram Alpha did not show ##\sum_{i=1}^n 1/(n+i) = \Psi (n+i-1) - \Psi (i+1)##. WA is smart enough to know that a summation over all ##i## from 1 to ##n## cannot contain ##i## in its answer!

    Anyway, getting the exact answer like that is not very useful. Another, much, much simpler method gets to the answer quickly and easily; unfortunately, the merest hint of how it works will immediately give away the whole answer.
     
  5. Aug 6, 2017 #4

    stevendaryl

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    That Wolfram result doesn't actually make sense, because there shouldn't be any [itex]i[/itex] in the result: [itex]i[/itex] is a dummy variable.

    Just playing around, we can write the original series in this way:

    [itex]\displaystyle \sum_{i=1}^n \frac{1}{n+i} = \displaystyle \sum_{i=n+1}^{2n} \frac{1}{i} = \displaystyle \sum_{i=1}^{2n} \frac{1}{i} - \displaystyle \sum_{i=1}^{n} \frac{1}{i}[/itex]

    The harmonic series is defined by: [itex]H_n = \displaystyle \sum_{i=1}^n \frac{1}{i}[/itex], so we can write the sum as:

    [itex]\displaystyle \sum_{i=1}^n \frac{1}{n+i} = H_{2n} - H_{n}[/itex]


    At this point, it seems that you could just use the asymptotic form of [itex]H_n[/itex] for large [itex]n[/itex].
     
  6. Aug 6, 2017 #5
    Very impressive view stevendaryl!!!

    I’m trying to use different approaches to see this tricky result. By what you see I could, at least, make an algorithm in C language to check if it really converges to [itex]log(2) = 0.69314718[/itex] :

    upload_2017-8-6_15-19-3.png
     
  7. Aug 6, 2017 #6

    Ray Vickson

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    If you look at it the right way it is not tricky at all, and can be done without any use of a computer, or calculator, or anything.

    Once you have assured me that you have turned in the work for marking, I will be happy to post the method. You will kick yourself once you see it!
     
  8. Aug 6, 2017 #7
    Actually I’m still trying to see a way to find out this result and I’m doing my best to research in different sources. But you can feel free to send the method as well as the Liu Kang to kick myself =)
     
  9. Aug 6, 2017 #8

    Ray Vickson

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    I cannot send the method as long as the work has not yet been turned for marking. That would be against the PF rules.
     
  10. Aug 6, 2017 #9

    LCKurtz

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    @Ray Vickson: I don't think giving the hint of thinking about an approximating sum to an integral is too much of a hint.
     
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