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Infinite Series & Convergence

  1. Mar 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine convergence for each of the following:

    ∑ 1 / [n (log n)2]

    ∑ 1 / [n log n log(log n)]

    [log=ln=natural log]

    2. Relevant equations

    3. The attempt at a solution
    I learned the root test, ratio test, comparsion test, and integral test. But how can we figure out what test to use? Would the "ratio test" work here?

    Any help is appreciated!
  2. jcsd
  3. Mar 12, 2010 #2
    I don't think the ratio test would help since you'll end up with ln(n + 1).
    Try comparing n*(ln(n))2 with something that grows slower for x > 2.
  4. Mar 12, 2010 #3
    hmm...Why wouldn't the ratio test work?? What's wrong with having ln(n+1)??

    I'm just confused becuase I don't know which test to use when I look at those problems...

    Should I use the integral test? Why or why not?

  5. Mar 12, 2010 #4
    Yes, you can use the integral test (I should go to bed :blushing: too tired)

    There's no way to simplify ln(n + 1). I didn't look much at doing it with the ratio test; it might be possible, but I think it would be harder than any other way.
  6. Mar 12, 2010 #5


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    You just have to try each test out and see if it'll work out. Eventually, you'll develop a feel for which test is probably the best for a series.
  7. Mar 12, 2010 #6
    I have some problem applying the integral test for the second sum...

    Integral test: if f is continuous, decreasing, and POSITIVE on [2,∞), then

    ∑ f(k) converges if and only if

    ∫ f(x) dx converges.

    f(x) = 1 / [x log x log(log x)]
    But f is not positive on [2,∞)
    e.g. f(2)<0, f(2.1)<0

    So how can we apply the integral test??

    Please help...I'm confused...
  8. Mar 12, 2010 #7


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    Ok, then apply it on [3,infinity).
  9. Mar 12, 2010 #8
    But is that ok? Will it affect convergence/divergence?

    ∫ f(x) dx
    Does the value of "k" here have any effect on convergence/divergence?
  10. Mar 12, 2010 #9


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    You can drop any finite number of terms without affecting convergence or divergence of the series. So no, k doesn't really matter.
  11. Mar 12, 2010 #10
    For infinite series, the place where it starts has no effect on convergence/divergence.
    But is it also true for improper INTEGRALS?

    ∫ f(x) dx converges

    ∫ f(x) dx diverges
    Is this possible??
  12. Mar 12, 2010 #11


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    Sure, it's possible. That's why you have to check the premises of the integral test before you apply it. f(x) has to be nonnegative and monotone decreasing on the part of the series you want to apply it to. Just like in this case you should move k from 2 to 3.
  13. Mar 12, 2010 #12
    Integral test: if f is continuous(?), decreasing, and POSITIVE on [n,∞), then

    ∑ f(k) converges if and only if

    ∫ f(x) dx converges.

    To apply the integral test, does f have to be continuous? In my first year calculus text book, the integral test is as stated above; f is assumed to be continuous. But in my other textbook, there is no such assumption. Or maybe there is a typo?

    What is the correct statement of the integral test? Do we need to assume continuity?
  14. Mar 12, 2010 #13

    Char. Limit

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    Yes, f(x) has to be continuous. If there is a infinite discontinuity, it will seriously affect convergence. If there is a jump discontinuity, then your function isn't likely to be easily integrable.

    Of course, [tex]\frac{1}{x log(x) log(log(x))}[/tex] is quite easily integrable and is continuous from 3 to infinity. Go for it.
  15. Mar 13, 2010 #14
  16. Mar 13, 2010 #15

    Char. Limit

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    I believe (and I might be wrong) that an integral implies continuity.

    Maybe even right... have you tried integrating 1/x over an interval that includes 0?
  17. Mar 13, 2010 #16
    But I think it is continuity implies integrable, not vice versa...

    What can happen to the "Integral Test" if f is not continuous???

    (actually in my real analysis book, they didn't mention "continuity" in the integral test at all, weird...)
  18. Mar 13, 2010 #17


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    Think about why the integral test works. f(x) is decreasing. You can make part of your series into an upper sum for the integral and part into a lower sum. Draw a picture. You'll see continuity isn't necessary. Though I can't think of a single reason to take f(x) to be discontinuous.
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