1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding 1/n + 1/(n+1) + + 1/2n in terms of the Euler's constant

  1. Sep 12, 2009 #1
    1. The problem statement, all variables and given/known data

    By using the Euler's constant (0.577...), express the limit of the sum 1/n + 1/(n+1) + ... + 1/2n explicitly as n goes to infinity.

    2. Relevant equations

    The limit as n goes to infinity for [1/1 + 1/2 + ... + 1/n - In (n)] is the Euler constant, 0.577...

    It has been deduced and shown previously that the limit of the sum 1/n + 1/(n+1) + ... + 1/2n is between 0.5 and 1. And the value lies somewhere around 0.6-0.75.

    3. The attempt at a solution

    I think the word "explicitly" in the question means that we need to express the answer in terms of the Euler's constant.

    limit as n goes to infinity for [1/1 + 1/2 + ... + 1/n - In (n)] = 0.577...

    Let a (subscript k) = 1/k + 1/(k+1) + ... + 1/2k and k<n

    limit as k goes to infinity for [1/1 + 1/2 + ...1/(k-1) + 1/k + 1/(k+1)+...+1/2k +1/(2k+1) ... 1/n - In (n)] = 0.577...

    limit as k goes to infinity for [1/1 + 1/2 + ...1/(k-1) + a(subscript k) + 1/(2k+1) ... 1/n - In (n)] = 0.577...

    limit as k goes to infinity for [a(subscript k)] = 0.577... - limit as k goes to infinity [1/1 + 1/2 + ...1/(k-1)] - limit as k goes to infinity [(1/(2k+1) + ... + 1/n] + limit as k goes to infinity [In(n)]

    Then I don't know how to continue. Perhaps I need to define other variables? It seems that k and n are a bit confusing...

    Thanks.
     
  2. jcsd
  3. Sep 13, 2009 #2

    VietDao29

    User Avatar
    Homework Helper

    This is wrong!! What is n in your expression? Is n related somehow to k?

    ------------------------------

    Ok, here's how I would tackle the problem. Let's denote the Euler Constant to be [tex]\gamma[/tex].

    Let [tex]x_n = \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} - \ln(n)[/tex]

    We know that:

    [tex]\gamma = \lim_{n \rightarrow \infty} x_n = \lim_{n \rightarrow \infty} \left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} - \ln (n) \right)[/tex]

    Since the limit of the sequence xn exists, all of its sub-sequences should have the same limit, too, right? Now consider the sub-sequence x2n

    [tex]\lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{2n} - \ln (2n) \right) = \gamma[/tex]

    [tex]\Rightarrow \lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left[ {\color{red}\left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} \right)} + {\color{blue}\left( \frac{1}{n + 1} + ... + \frac{1}{2n} \right)} - \ln (2n) \right] = \gamma[/tex]

    You want to find the limit of the blue part, right?

    Now, look at the red one, do you realize anything? Does it look "nearly" the same as something that you've already known? What can you do to make use of the Euler's Constant?

    Well, let's see if you can find a way to continue. :)
     
  4. Sep 13, 2009 #3
    Oh, the subsequence concept, yup. That's useful.

    So, gamma = [lim as n goes to infinity (1/1+1/2+...+1/n)] + [lim as n goes to infinity (1/(n+1) + ... + 1/2n)] - [lim as n goes to infinity (In (2n))]

    Gamma = Gamma + lim as n goes to infinity (In (2n)) +[lim as n goes to infinity (1/(n+1) + ... + 1/2n)] - [lim as n goes to infinity (In (2n))]

    Canceling gamma at both sides, we have:
    [lim as n goes to infinity (1/(n+1) + ... + 1/2n)] = [lim as n goes to infinity (In (2n))] - lim as n goes to infinity (In (2n))

    [lim as n goes to infinity (1/(n+1) + ... + 1/2n)] = lim as n goes to infinity [In (2n/n)] = lim as n goes to infinity In (2) = In(2) = 0.693...

    I used my Texas Instrument to calculate the value. When n = 71, the value obtained is 0.703723..., which is very close to In(2), so I believe In 2 is the correct answer?

    Thanks for the help. ;)
     
  5. Sep 13, 2009 #4

    VietDao29

    User Avatar
    Homework Helper

    Well, you're very close to correct.. :rolleyes:

    Remember that, you can only break the limit of 2 sums into the sum of 2 limit, when both of the 2 limits exist:

    [tex]\lim (x_n + y_n) = \lim x_n + \lim y_n[/tex] only holds if the limits of both sequences xn, and yn exist.

    Both expressions diverge, so you cannot split it like that. try this instead:

    [tex]\Rightarrow \lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left[ {\color{red}\left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n} -\ln (n) \right)} + \left( {\color{blue}\left( \frac{1}{n + 1} + ... + \frac{1}{2n} \right)} - \ln (2n) + \ln(n) \right) \right] = \gamma[/tex]
     
  6. Sep 13, 2009 #5
    Hm, everything looks good, but I think In 2 is the limit of 1/(n+1) + ... + 1/2n. The question asks for 1/n + 1/(n+1) + ... + 1/2n, so the 1/n term is missing there. ;( But In 2 is correct I think.
     
  7. Sep 13, 2009 #6

    VietDao29

    User Avatar
    Homework Helper

    You can modify it a little bit. It's still the same. How about, putting the term 1/n into the blue parentheses, and modifying the ln(...) terms a little bit? :)
     
  8. Sep 13, 2009 #7

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    Even better, try this instead:
    [tex]\Rightarrow \lim_{n \rightarrow \infty} x_{2n} = \lim_{n \rightarrow \infty} \left[ {\color{red}\left( \frac{1}{1} + \frac{1}{2} + ... + \frac{1}{n-1} -\ln (n-1) \right)} + \left( {\color{blue}\left( \frac{1}{n} + \frac{1}{n + 1} + ... + \frac{1}{2n} \right)} - \ln (2n) + \ln(n-1) \right) \right] = \gamma[/tex]
     
  9. Sep 13, 2009 #8
    Yay, this is great.
    So, lim [1/1 + 1/2 + ... + 1/(n-1) - In(n-1)] = gamma

    So the equation can be reduced to lim (1/n + 1/(n+1) + ... + 1/(2n)) = lim In[2n/(n-1)] = lim In 2 = In 2. Is this correct?

    Thanks both of you for the help. ;)
     
  10. Sep 13, 2009 #9

    VietDao29

    User Avatar
    Homework Helper

    Looks fine to me. :)
     
  11. Sep 13, 2009 #10
    I am so happy to finally solve the problem. ;) Thanks again and have a nice day!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding 1/n + 1/(n+1) + + 1/2n in terms of the Euler's constant
  1. Problem 2n-1<n! (Replies: 26)

Loading...