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Simplifying a series with the ratio test

  1. Feb 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine if the following series is divergent or convergent:

    ## ∑_1^∞ \frac {(2)(4)(6)...(2n)}{n!} ##


    2. The attempt at a solution

    I understand this can be simplified to:

    ## ∑_1^∞ \frac {(2^n)(n!)}{n!} ##
    This can easily be seen to be divergent.

    But when I do ## lim_{n→∞}\frac {a_{n+1}}{a_{n}} = lim_{n→∞}\frac {1}{n} ## after simplifying. Does anyone know what I may be doing wrong or why this is not appropriate? Is there any restriction on when the ratio test can be used? Thank you! :)
     
  2. jcsd
  3. Feb 24, 2014 #2

    AlephZero

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    How did you get that? (It is wrong).
     
  4. Feb 24, 2014 #3
    I know ## a_n = \frac {(2)(4)(6)...(2n)}{n!} ##
    So ##\frac {a_{n+1}}{a_n} = [\frac {(2)(4)(6)...(2(n+1))}{(n+1)!}] [\frac {n!}{(2)(4)(6)...(2n)}]##
    This simplifies to: ## \frac {2(n+1)}{2n} \frac {1}{n+1}##

    Now, when the limit is taken as n→∞, then ## \frac {2(n+1)}{2n} = 1## and you're left with ##lim_{n→∞}\frac {1}{n+1} ## I just simplified this to ##\frac {1}{n}##. Does anyone mind telling me what I did wrong here?
     
  5. Feb 24, 2014 #4

    Dick

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    No, the ratio is 2(n+1)/(n+1)=2. Where did the extra n come from??
     
  6. Feb 24, 2014 #5
    ahh i forgot to cancel a 2n
     
  7. Feb 25, 2014 #6

    Ray Vickson

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    Why bother with the ratio test? The nth term of the series equals ##2^n## (as you have already noted). Since ##2^n## does not go to 0 as ##n \to \infty## the series cannot possibly converge.
     
  8. Feb 26, 2014 #7
    Yep, I just didn't know why using a different test that I wasn't getting the same solution. Problem's solved.
     
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