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Comparison test for convergence problem: why is this incorrect?

  1. Feb 27, 2012 #1
    1. The problem statement, all variables and given/known data

    The original question is posted on my online-assignment. It asks the following:

    Determine whether the following series converges or diverges:

    [itex]\sum^{\infty}_{n=1}[/itex][itex]\frac{3^{n}}{3+7^{n}}[/itex]

    There are 3 entry fields for this question. One right next to the series above with the following options:

    either [itex]\succ[/itex] or [itex]\prec[/itex] then next to that there is a field in which to input the thing that I am going to compare it to.

    the third and final field I choose divergent or convergent

    2. Relevant equations



    3. The attempt at a solution

    So I compared it to ([itex]\frac{4^{n}}{6^{n}}[/itex]) because the original series is clearly less than this one. By doing the comparison test I determined that the series converges.

    So I get "less than" right and "converges" right but I didn't get the other part right. Isn't it true that the original series is less than the one I decided above? Or was I wrong somewhere else?
     
  2. jcsd
  3. Feb 27, 2012 #2

    lanedance

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    Everything you said sounds trues, though I'm not sure I understand the question?

    Also, why wouldn't you just compare to [itex] \sum_n \frac{3^n}{7^n} [/itex]?
     
    Last edited: Feb 27, 2012
  4. Feb 27, 2012 #3
    That must be it.. I was just slight confused as to whether it was bigger or smaller than the original series, so I wanted to be EXTRA sure by choosing the one I mentioned above. I guess that must have been where I went wrong. But still, is what I did above correct? I realize that I could have made an "easier" decision for what to compare it to, but isn't what I chose still correct?

    Thanks!
     
  5. Feb 27, 2012 #4

    lanedance

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    yeah looks ok to me, generally want to choose the easiest to compare and the closest possible.

    For the pupose of the convergence, as long as for some n>N each term yn> xn and (sum yn) converges, then (sum xn) converges
     
  6. Feb 27, 2012 #5

    kai_sikorski

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    Gold Member

    If it's an automatically graded online assignment the software might be hard-coded to only accept the most natural choice. What you did is correct though.
     
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