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Given general term - find limits and comparison

  1. Nov 20, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##\displaystyle a_n=\frac 1 2+\frac 1 3+....+\frac 1 n##. Then

    A)##a_n## is less than ##\displaystyle \int_2^n\frac{dx}{x}##.

    B)##a_n## is greater than ##\displaystyle \int_1^n\frac{dx}{x}##.

    C)##\displaystyle \lim_{n\rightarrow \infty} \frac{a_n}{\ln n}=1##

    D)##\displaystyle \lim_{n\rightarrow \infty} a_n## is finite.

    2. Relevant equations



    3. The attempt at a solution
    To my knowledge, there is no known closed form for the given ##a_n##.

    I am clueless about the right approach so I started with ##n=2##. For n=2, ##a_2=0.5##

    Also,
    $$\int_2^2 \frac{dx}{x}=0$$
    and
    $$\int_1^2 \frac{dx}{x}=\ln 2 \approx 0.693$$
    Obviously, A and B are not the answers.

    How do I check for other options?

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Nov 20, 2013 #2
    D is not correct the series is diverging. C is basically saying that a(n) = ln(n) if n becomes large is that correct?
     
  4. Nov 20, 2013 #3
    I don't have the answers at the moment, I will have them by tomorrow.

    Can you please explain how do you get the series to be diverging?

    Thanks!
     
  5. Nov 20, 2013 #4

    LCKurtz

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    Think about comparing the series with approximating sums for the integrals in A and B.
     
  6. Nov 20, 2013 #5
    I am not sure if I understand your statement but do you ask me this:
    $$a_n=\int_2^n \frac{dx}{x}$$
    ?
     
  7. Nov 20, 2013 #6

    LCKurtz

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    Yes. Think about approximating that with rectangles and see if you can relate it to your series.
     
  8. Nov 20, 2013 #7
    Thanks LCKrutz! :)

    I just found the wiki page on the given series titled "Harmonic Number" and it contains a nice sketch of approximation using rectangles. http://en.wikipedia.org/wiki/Harmonic_number

    Getting back to the question, as ##n\rightarrow \infty##,
    $$a_n=\ln n-\ln 2$$
    ##a_n## is obviously not finite. That leaves us with option C.

    $$\lim_{n\rightarrow \infty} \frac{a_n}{\ln n}=\lim_{n\rightarrow \infty} \frac{\ln n-\ln 2}{\ln n}=\lim_{n\rightarrow \infty} 1-\frac{\ln 2}{\ln n}=1$$
    Hence, C is correct, thanks a lot LCKurtz! :smile:
     
  9. Nov 20, 2013 #8

    LCKurtz

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    What is the exact statement of the problem? Is it a True-False type question?

    [Edit] I see you answered that question while I was posting it.
     
  10. Nov 20, 2013 #9
    What I wrote is the exact wording of the problem statement but I should have been more clear. The problem is from a test paper belonging to the section which consists of multiple choice questions. I have to select the correct options.
     
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