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Convergent/Divergent Series

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    I have to determine whether the given series is convergent or divergent using the comparison tests:
    [tex]\sum[/tex] from n = 1 to infinity of (n + 4n / (n + 6n)


    2. Relevant equations
    If bn is convergent and an [tex]\leq[/tex] bn then an is also convergent

    liimit of an/bn as n goes to infinity = c, if c > 0, then both are either convergent or divergent

    3. The attempt at a solution

    I tried saying that bn was (4/6)^n but i dont know how to compare that to the original series
     
  2. jcsd
  3. Mar 30, 2009 #2

    HallsofIvy

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    Certainly [itex](n+4^n)/(n+ 6^n)< (n+ 4^n)/6^n[/itex] because the left side has a larger denominator. It is also true that n< 4^n for any positive integer n. That means that [itex]n+ 4^n< 4^n+ 4^n< 2(4^n)[/itex] and so [itex](n+4^n)/(n+6^n)< (n+4^n)/6^n< 2(4^n)/6^n)[/itex].
     
  4. Mar 30, 2009 #3
    so then because 2(4/6)^n is convergent, the original series is convergent, correct?
     
  5. Mar 30, 2009 #4
    Correct.
     
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