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Using the ratio test to figure if the series is convergent

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\sum[/itex] ftom n=1 to [itex]\infty[/itex] (-2)n/nn.

    3. The attempt at a solution
    limn->[itex]\infty[/itex] | (-2)n+1/(n+1)n+1) x nn/(-2)n | = |-2|limn->[itex]\infty[/itex] |(n/n+1)n*(1/n+1) |

    If it were only (n/n+1) then would the answer be 2e? Either way, how do you sole this the way it is?
  2. jcsd
  3. Mar 8, 2013 #2


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    I think there are brackets missing.
    (n/(n+1)^n) is smaller than 1, that is sufficient here. The other factor is more interesting.
  4. Mar 8, 2013 #3
    Another way to look at it: The growth rate of n^n is faster than that of n!. If you replace n^n with n!, where does that leave you at?
  5. Mar 8, 2013 #4

    Ray Vickson

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    If n > 3 then (2/n)^n < (2/3)^n.
  6. Mar 8, 2013 #5


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    Notice that by the ratio test, the series of positive terms will converge and so Ʃan will be absolutely convergent which implies Ʃan will converge.
  7. Mar 9, 2013 #6
    By the ratio test i'm left with: 2 x limx->[itex]\infty[/itex] |nn/(n+1)n+1)| which is the same as what i previously wrote: 2 x limx->[itex]\infty[/itex] [n/(n+1)]n*1/(n+1).

    Dividing by the largest power of the polynomial in the denominator:
    2 x limx->[itex]\infty[/itex] [1/(1+1/n)]n*1/(n+1).-
    so it becomes: 2 x limx->[itex]\infty[/itex] |1n*[1/(1+n)] = 2 x limx->[itex]\infty[/itex] 1n*0=0
    ratio test = 0>1 therefore the series is absolutely convergent and therefore convergent.

    The only thing im not sure about doing in this problem really is how to deal with the numbers when i divide them by n when i'm working out the limit.
  8. Mar 9, 2013 #7


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    You can always divide numerator and denominator by n, if you do it consistently (as n!=0).
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