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Showing convergence

  1. Jun 9, 2010 #1
    1. The problem statement, all variables and given/known data
    (ln(n))^2/sqrt(n)


    2. Relevant equations
    Various series tests


    3. The attempt at a solution
    The answer showed that the series converges to 0, I know I can rule out the nth term test because the limit is 0. However, I'm not sure how to go about this since it's a logarithmic problem. Can someone direct me as to which test to use to get to thsi? Maybe comparison?
     
  2. jcsd
  3. Jun 9, 2010 #2
    Think about comparison test.

    *Good idea to use latex: [tex] \sum_{n=1}^\infty \frac{\ln^2{n}} {\sqrt{n}}[/tex]
     
    Last edited: Jun 9, 2010
  4. Jun 9, 2010 #3

    lanedance

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    is this a series or a sequence?
     
  5. Jun 9, 2010 #4
    its a sequence
     
  6. Jun 9, 2010 #5
    what exactly should i compare it to, (im not good with natural logs) I was thinking maybe 1/sqrt(n) but that doesn't show it convergest to 0
     
  7. Jun 9, 2010 #6
    1. What the comparison test says?

    2. What is the relation between [tex] \frac {1} { \sqrt{x} }\ \ and\ \ \frac{\ln^2{x}}{\sqrt{x}}\ ? [/tex]
     
  8. Jun 9, 2010 #7

    lanedance

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    if the sequence is:
    [itex] a_n = \frac{ln(n)^2}{\sqrt{n}} [/itex]

    how about seeing whether you could show, for some n>N that:
    [itex] ln(n)<n^{1/4} [/itex]
     
    Last edited: Jun 9, 2010
  9. Jun 9, 2010 #8
    wait...OHHH to prove convergence if its a series, u use a series test, but this is just a sequence!!! that means i just need to take the limit to infinity and if its a finite number it should converge!!! :D am i right?
     
  10. Jun 9, 2010 #9
    You converge to a solution...=)
    Look carefully what the comparison test means and says.
     
  11. Jun 9, 2010 #10
    if i recall, comparison test is when you compare a sequence that you don't know to a sequence that u do know converges/diverges
     
  12. Jun 9, 2010 #11
    Last edited: Jun 9, 2010
  13. Jun 9, 2010 #12

    lanedance

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    that stuff is for series, so whether the sum converges, say
    [itex]S_n = \sum_n a+n = \sum_n \frac{ln(n)^2}{\sqrt{n}} [/itex]

    if this is actually just a sequence of a_n, then showing a limit exists is sufficient
    [itex] \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{ln(n)^2}{\sqrt{n}} [/itex]

    vipertongn - it think you should make sure which it is... in this case the sequence converges to zero, however the series diverges
     
  14. Jun 9, 2010 #13
    it is a sequence, what confused me whas the fact that i find the series to diverge badly but the solutoins said it converged to zero :3 I understand it now
     
  15. Jun 9, 2010 #14
    [tex] \frac {1} { \sqrt{n} }<\frac{\ln^2{n}}{\sqrt{n}},\ \forall\ n>e [/tex]

    [tex] \sum_{n=e}^\infty \frac{1}{\sqrt{n}}\ is\ divergent\ so\ from\ the\ comparison\ test\ \sum_{n=e}^\infty \frac{\ln^2{n}}{\sqrt{n}}\ is\ also\ divergent\ [/tex]

    Both sequences converge to 0, but don't confuse the convergence of sequences to convergence of their series.
    It seems to me you're not familiar enough with the theory.
     
    Last edited: Jun 9, 2010
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