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Use comparison test to see if series converges

  1. Jan 21, 2017 #1
    1. The problem statement, all variables and given/known data

    [tex] \sum_{x=2}^{\infty } \frac{1}{(lnx)^9} [/tex]



    2. Relevant equations


    3. The attempt at a solution
    [tex] x \geqslant 2 [/tex]

    [tex] 0 \leqslant lnx < x [/tex]
    [tex] 0 < \frac{1}{x} < \frac{1}{lnx} [/tex]

    From this we know that 1 / lnx diverges and I wanted to use this fact to show that 1 / [(lnx) ^ 9] diverges but at k >= 2 we have this:

    [tex] 0 < \frac{1}{(lnx)^9} < \frac{1}{lnx} [/tex]

    So it doesn't really work.

    Is there a relation between 1/x and 1/[(lnx)^9] that I can use to solve the problem?

    Any help would be appreciated, thanks!
     
  2. jcsd
  3. Jan 21, 2017 #2

    PeroK

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    Hint: work out how to show the condition on ##x## such that ##\ln x < x## and see whether you can extend this for ##(\ln x)^9##.
     
  4. Jan 21, 2017 #3
    Well the condition on x to make that true is x > 1 right?

    To extend it, should I try and find when this is true:

    [tex] (lnx)^9 < x [/tex]

    ?
     
  5. Jan 21, 2017 #4

    PeroK

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    How do you prove that?
     
  6. Jan 21, 2017 #5
    Well, lnx <= 0 whenever 0 < x <= 1 so that bit is obvious I guess.

    Also, whenever x > 1, lnx < x because it's a monotonically increasing function with first derivative:

    1 - 1/x > 0.

    For that to be true x > 1.
     
  7. Jan 21, 2017 #6

    PeroK

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    Okay. But can you extend that idea to ##(\ln x)^2##?

    Or, is there another way to do it that might be easier to extend?
     
  8. Jan 21, 2017 #7
    I don't think you can because if you take lnx where x < 1, it's a negative result. So (lnx)^2 would be positive.

    Another way to prove lnx < x?
     
  9. Jan 21, 2017 #8

    PeroK

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    Yes, just think of what you could do to that equation.
     
  10. Jan 21, 2017 #9
    Oh right,

    [tex] lnx < \sqrt{x} [/tex]
    [tex] \sqrt{x} - lnx > 0 [/tex]
    [tex] \frac{\mathrm{d} }{\mathrm{d} x}(\sqrt{x}-lnx)=\frac{1}{2\sqrt{x}}-\frac{1}{x} > 0 [/tex]
    when x >= 2.

    so if that's true then so is this:
    [tex] (lnx)^2 < x [/tex]

    Is that right?
     
  11. Jan 21, 2017 #10

    PeroK

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    Let me help. It seemed obvious to me:

    ## \ln x < x## iff ##x < e^x##

    Perhaps that's easier to extend to powers of the logarithm.
     
  12. Jan 21, 2017 #11
    Ok, so we can say that ## \ln x < e^x## so ## (\ln x)^9 < e^9x ## and ## 1/e^9x < 1/(lnx)^9##

    Is that the idea? (those x's should be in the exponent as well)
     
  13. Jan 21, 2017 #12

    PeroK

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    That maths is all awry!

    Let me help you a bit more:

    Generally, you can use the following result without proof:

    For all ##n## eventually ##e^x## is greater than ##x^n##. In fact the limit ##\frac{e^x}{x^n}## is ##+\infty## as ##x \rightarrow +\infty##

    That result should help in this case.
     
  14. Jan 21, 2017 #13
    Ok, yeah I got no idea.

    I'm pretty sure 1/x < 1/(lnx)^9, I just can't prove it like I can for 1/x < 1/lnx.

    And since 1/x diverges so does 1/(lnx)^9.

    I'm not sure how the facts that lnx < x iff x < e^x (is this ever false?) and limit of e^x / x^n -> infinity helps.
     
  15. Jan 21, 2017 #14

    PeroK

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    You know that ##\ln## is the inverse of ##\exp##?
     
  16. Jan 21, 2017 #15
    Yes
     
  17. Jan 21, 2017 #16

    PeroK

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    Well do something with that fact!
     
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