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How to find limit of arithmetic series?

  1. Feb 12, 2008 #1
    Find the limit of the following series:

    lim (n-->infinity) 1 + 3 + 5 + 7 + ... (2n-1) / (n+1) - [(2n+1)/2]




    3. From what I kno this is an arithmetic series, meaning I must use that arithmetic series formula. so its (first term + last term / 2 times the number of terms) n^2..

    now my teacher did another step, and this is where im lost:

    n^2/n+1 - (2n+1)/n

    I can solve it from here but can someone be kind enough to explain the above part to me. (by the way, answer is -3/2)


    I also have another question, do i need to divide by highest power of n (in this case) of denominator to all numbers when n approaches infinity? So i need to do that everytime?

    Any comments are most appreciated and I thank you for your time :-)
     
    Last edited: Feb 12, 2008
  2. jcsd
  3. Feb 13, 2008 #2

    Gib Z

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    Homework Helper

    Can you post your question again, with more brackets :( Its hard to tell what you mean.

    In any case, [tex]1+3+5...+ n = \left( \frac{n+1}{2}\right)^2[/tex].

    And when you have the quotient of polynomials, to find the limit always divide through by the highest power of x.
     
  4. Feb 13, 2008 #3
    yes okay sorry

    [1 + 3 + 5 ...(2n-1)] / (n+1) - [(2n+1)/2]

    so 1 is the first term, and 2n-1 is the last term of the series. I don't know what the other part after the subtraction sign is...

    i hope this helped
     
  5. Feb 13, 2008 #4

    HallsofIvy

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    No, this is not an arithmetic series! An arithmetic series, where the numbers are getting larger and larger, can't possibly converge. Here you might have an arithmetic series divided by something. It is not clear whether you mean
    [tex]\frac{1+ 3+ 5+ 7+ \cdot\cdot\cdot+ (2n-1)}{(n+1) - \frac{2n+1}{2}}[/tex]
    or
    [tex]\frac{1+ 3+ 5+ 7+ \cdot\cdot\cdot+ (2n-1)}{(n+1)}- \frac{2n+1}{2}[/tex]

    Unfortunately, I can't see that either of those converges- and, in fact, what your teacher gives does not converge!

    Well, you don't NEED to- if you can find some other way to do the limit. But it certainly is usually the easiest way to do such a limit. The obvious point is that it is easier to work with "0" than with infinity! Dividing each term by the highest power puts n into the denominator of each individual fraction- as n goes to infinity, each fraction goes to 0.
     
    Last edited by a moderator: Feb 13, 2008
  6. Feb 13, 2008 #5
    ye the whole series is divided by n+1..

    edit: nvm i solved it...
     
    Last edited: Feb 13, 2008
  7. May 4, 2008 #6
    the above poster didn't say it, so i will because it just helped me while doing my calculus homework here:

    thanks
     
  8. Oct 21, 2011 #7
    lim (n→∞) 1 + 3 + 5 + 7 + ... (2n-1) / (n+1) - [(2n+1)/2]

    first we find arithmetic sum: Sn=n/2*[2a1+(n-1)*d], where d=2, a1=1

    lim (n→∞) n/2*[2+(n-1)*2] / (n+1) - [(2n+1)/2]

    lim (n→∞) n/2*[2+2n-2] / (n+1) - [(2n+1)/2]

    lim (n→∞) n(n) / (2n-1) / (n+1) - [(2n+1)/2]

    lim (n→∞) n^2/ 2(n+1)-2n+1/2

    lim(n→∞)n^2/ 2n+2-2n+1/2

    lim(n→∞)n^2/ 2/3

    2/0

    "infinity" is the result
     
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