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Comparison test

  1. Feb 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Using the limit comparison test to determine convergence or divergence


    (1) [itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{3}}{n^{3}}[/itex]


    (2) [itex]\sum^{\infty}_{n=1} \frac{1}{\sqrt{n} ln(x)}[/itex]


    (3)[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{2}}{n^{\frac{3}{2}}}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    I have problem find thing the another series (Bn) compare with An. First I choose Bn when 'n' grows large to represent An

    The first question I chose Bn = [itex]\frac{1}{n^{3}}[/itex]
    it is the p-series and p>1 so it converges

    and then
    [itex]lim_{n\rightarrow\infty} \frac{\frac{ln(n)^{3}}{n^{3}}}{\frac{1}{n^{3}}}[/itex]

    [itex]lim_{n\rightarrow\infty} ln(n)^{3} = \infty[/itex]

    which is wrong so I change to choose Bn = [itex]\frac{1}{n^{2}}[/itex]

    [itex]lim_{n\rightarrow\infty} \frac{\frac{ln(n)^{3}}{n^{3}}}{\frac{1}{n^{2}}}[/itex]

    [itex]lim_{n\rightarrow\infty} \frac{ln(n)^{3}}{n}= 0[/itex]


    So I can conclude that An is convergent by limit comparison

    but why I cannot use Bn = [itex]\frac{1}{n^{3}}[/itex] ????

    =========================================

    (2) and (3) are the same as my first question

    (2) [itex]\sum^{\infty}_{n=1} \frac{1}{\sqrt{n} ln(x)}[/itex]

    I chose Bn = [itex]\frac{1}{n^{1/2}}[/itex] but after I took the limit of An/Bn. I got 0. In the answer shows that I have to choose Bn = [itex]\frac{1}{n^{}}[/itex]

    (3)[itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{2}}{n^{\frac{3}{2}}}[/itex]

    I chose Bn = [itex]\frac{1}{n^{3/2}}[/itex] after I took the limit of An/Bn. I got [itex]\infty[/itex]. In the answer shows that I have to choose Bn = [itex]\frac{1}{n^{5/4}}[/itex]

    Please give some comments and advices

    Thank you
     
    Last edited by a moderator: Feb 10, 2013
  2. jcsd
  3. Feb 10, 2013 #2

    mfb

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    Staff: Mentor

    The idea is to compare your series with a bigger one with known convergence. 1/n^3 is not bigger, therefore it does not work.

    The same happens in the other direction, if you want to show divergence.
     
  4. Feb 10, 2013 #3
    Thank you for the comments


    [itex]\sum^{\infty}_{n=1} \frac{(ln(n))^{3}}{n^{3}}[/itex]

    The series (Bn) that we choose must represent the original series (An) and we recognize converges or diverges from Bn. (We dont know from the original). the problem is...

    How do I know that [itex] \frac{1}{n^3} [/itex]is too small [itex]\frac{(ln(n))^{3}}{n^{3}}[/itex] and[itex] \frac{1}{n^2} [/itex]not too big for [itex]\frac{(ln(n))^{3}}{n^{3}}[/itex] and why not choose[itex] \frac{1}{n}[/itex]. still not get it.
     
  5. Feb 10, 2013 #4

    mfb

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    2016 Award

    Staff: Mentor

    1/n does not give a converging series.
    You could choose 1/n3/2 or 1/n5/2 or something similar, if you like, but 1/n2 is easier to study.
     
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