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Convergent or divergent?

  1. Mar 22, 2009 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series converges or diverges.

    [tex]\sum[/tex] 3+7n / 6n




    Attempt :

    Comparison test :

    3+7n / 6n < 7n / 6n

    3+7n / 6n < (6/7)n

    since (6/7)n is a geometric series and is convergent is
    3+7n / 6n convergent as well?
     
  2. jcsd
  3. Mar 22, 2009 #2
    I see my mistake

    3+7n / 6n > (7/6)^n

    but then what?
     
  4. Mar 22, 2009 #3
    Hint: For a series
    [tex]\sum_{n=0}^{\infty}a_n[/tex]

    to converge, the terms have to converge to zero, i.e.

    [tex]\lim_{n->\infty}a_n=0[/tex].
     
  5. Mar 22, 2009 #4
    I could use the limit comparison test but I get stuck.
     
  6. Mar 22, 2009 #5
    You have [tex]a_n=3+7^n/6^n>3[/tex], so can [tex]a_n[/tex] converge to zero?
     
  7. Mar 22, 2009 #6
    How did you figure that inequality ?

    (7/6)^n converges to -7
     
  8. Mar 22, 2009 #7
    Clearly, (7/6)^n is a positive number for any n.

    I don't think so, 7/6>1, so (7/6)^n goes to infinity for large n.
     
  9. Mar 22, 2009 #8
    (7/6)^n

    This is a geometric series. And I know that by definition a*r^(n-1) ,where |r|<1 = a/(1-r).

    so (7/6)^n
    =

    (7/6) * (7/6)^(n-1)

    so,
    a = 7/6
    r = 7/6

    it follows that

    (7/6)^n = a/(1-r) = (7/6) / (1-7/6) = -7
     
  10. Mar 22, 2009 #9
    wait I see what your saying. How comes this the above statement is wrong?
     
  11. Mar 22, 2009 #10
    no your right, 7/6 > 1 so this series diverges.
    Thanks
     
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