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Math Help - Testing Series for Convergence/Divergence (2 Questions)

  1. #1
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    Testing Series for Convergence/Divergence (2 Questions)

    Hello All! Two series questions from my homework that I didn't know how to do:

    Objective: Test for Convergence or Divergence

    1) Sum (n=1 to Infinity) of 3n+2/(n(n+1))

    3n+2/(n^2+n). Since it's decreasing I can apply the Integral test, correct? I could split it into 3n/(n^2+n) = 3/n+1 (easy to integrate) and 2/(n^2 +n). I don't know how to integrate this second part -- unless you can do it by partial fractions with A/(x) + B/(x+1). That might work actually. Is this correct?

    2) Sum (n = 2 to Infinity) of 1/(n*(ln(n)^2). With the integral test (f(x) = 1/(x*(ln(x)^2), I could set u = ln(x), and du = 1/x, to get lim (t->Infinity) of Integral u^2du, but the bounds would be strange right? They would be ln(2) and ln(t). I could carry on this way but is this correct? I hate setting up a problem incorrectly and then having to work all the way through it!

    Thanks once again: this forum is fantastic!!!!
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  2. #2
    MHF Contributor Calculus26's Avatar
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    For the first if either diverges the whole thing diverges

    So yopu only have to consider 3/(n+1)
    You could of course do a comparison

    (3n+2)/(n^2 +n) > 3n/(n^2 +n) = 3/(n+1) > 3/(2n)= (3/2)(1/n)

    which is your divergent harmonic sereies

    For #2 is ln(n)^2 in the numerator ?

    you can always work out the AD first and use the original limits

    but still no problem as you'll end up with [ (ln(ln(t))]^3 which -> infinity

    as t->inf
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  3. #3
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    For #2, ln(n)^2 is in the denominator. If i say u = lnx, du = 1/x, lim (t-> infinity) of Integral from ln(2) to ln(t) of 1/u^2.

    Results in -1/u (ln(2) to ln(t) results in -1/(ln(t)) + 1/(ln(2)) as t --> Infinity. -1/ln(t) approaches 0, so integral = 1/(ln(2)) correct? So convergent?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    You've got it !
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