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Please help quick, calc help (limits)

  1. Nov 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Two questions

    lim as n -> infinity

    (1^k + 2^k + ... + n^k) / n^(k+1)

    and


    lim as n -> infinity


    1/(n+1) + 1/(n+2) + ... + 1/2n


    2. Relevant equations
    Definitions of limits, laws of exponents etc.


    3. The attempt at a solution

    Well I think I have them both solved but it seems too easy therefore I think I did something wrong or am missing something.

    I think they are both zero.

    The first one you can rewrite the term in the limit as

    (1^k)/(n^(k+1)) + ... + (n^k)/(n^(k+1))

    If you take the limit as n goes to infinity of each of these, then they all go to zero.

    I am not quite sure I can do this, or argue this that way, but it is the only thing that comes to mind.

    For the second question, I do the same thing. Take the limit as n goes to infinity of each term. Since they all go to 0 the entire limit goes to zero.


    Please let me know if my reasoning is correct.
     
  2. jcsd
  3. Nov 12, 2008 #2

    berkeman

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    Staff: Mentor

    One quick santiy check -- use Excel to do a few examples of each summation...

    Start with n=10, then cut and paste to make n=20, etc. Do you see a trend?
     
  4. Nov 12, 2008 #3

    berkeman

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    Staff: Mentor

    BTW, you will need a column for n, and a column for the numerator of the nth term, and a column for the denominator of the nth term, and a column for the quotient of the nth term, and then a sum over that last column down at the bottom.
     
  5. Nov 12, 2008 #4
    Yea they don't look to be zero, the first one looks like 1/3 and the second one looks like... well I don't know with n = 200 it was like .69 and n = 300 it was over 1.

    Can you please give me a hint on how to actually solve these guys?
     
  6. Nov 12, 2008 #5

    Dick

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    Science Advisor
    Homework Helper

    You must have taken k=2 for the first one, right? They are both Riemann sums which become exact in the limit as n -> infinity. The first one is the integral from 0 to 1 of x^k. The second one is the integral from 1 to 2 of 1/x. The first one depends on k, but if k=2 it's 1/3. That's how I knew you used k=2. The second is ln(2)=0.693... Good job using Excel (yuck!). Very brave. Review Riemann sums and see how this works, ok?
     
  7. Nov 12, 2008 #6
    Dick thank you very much for your response. I owe ya one, how about next time you need Excel help I'm your guy!
     
  8. Nov 12, 2008 #7

    Dick

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    Science Advisor
    Homework Helper

    Actually the once or twice in my life I've actually used a spreadsheet, I used Openoffice Calc. But I'm sure you can you can handle that as well. I'll keep you in mind.
     
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