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Limit comparison test.

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data

    heres the equations: http://img407.imageshack.us/img407/738/untitledzk8.jpg

    there are two equations, a and b.

    2. Relevant equations

    3/n^2 for a,

    and 1/n for b.

    3. The attempt at a solution

    Using the limit comparison test, i don't understand why to use 3/n^2 and 1/n for these equations.

    It was explained that we use the largest n in the numerator and denominator, but what we picked, clearly isn't the largest???

    please help. thanks.

    also, how does a converge, and b diverge??
    Last edited: Jan 25, 2007
  2. jcsd
  3. Jan 26, 2007 #2

    Gib Z

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    Homework Helper

    We use 3/n^2 because its a good comparison, a is similar to it. Same for b, if you dont include the 19n, it simplifies to 1/n.

    Now, 3/n^2 fulfils all the requirements to converge. The terms converge to zero etc. It can be shown that it is equal to [itex]\frac{\pi^2}{2}[/itex], thought I wont post the proof here.

    1/n is the harmonic series, represented by zeta(1) or a p series of p=1.

    A p series is of the form [tex]\sum_{n=1}^{\inf} \frac{1}{n^p}[/tex]. The rule of convergence of these is they converge if p>1, or diverge < or =1.
  4. Jan 26, 2007 #3


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    Staff Emeritus
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    For large values of n, the highest power of n will "dominate" (100000003 if far larger than 1000000002 [exactly 100000000 times as large!]) so we can ignore lower powers: In a, the numerator is "dominated" by 3n while the denominator is "dominated" by n3. For very large n, the fraction will be close to [tex]\frac{3n}{n^3}= \frac{3}{n^2}[/tex].

    In b, n2+ 19n is "dominated by n2 and the square root of that is n. For very large n, [tex]\frac{1}{\sqrt{n^2+ 19n}}[/itex] will be close to [tex]\frac{1}{n}[/tex].

    As for why one converges and the other diverges, use the "integral test":
    [tex]\int \frac{3}{x^2}dx= -\frac{3}{x}[/tex]
    which goes to 0 as x goes to infinity.
    On the other hand,
    [tex]\int \frac{1}{x}dx= ln(x)[/itex]
    which does not converge as x goes to infinity.
    Last edited: Jan 26, 2007
  5. Jan 26, 2007 #4
    awesome job, halls thanks for showing the simplification too. i see it now. thanks.
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