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L.C.T. Misgraded Cal. Problem?

  1. Dec 8, 2004 #1
    L.C.T. Misgraded Cal. Problem!???


    I just took a Cal. II test and used the Limit comparison test on one of my series. Here it is:

    / n^2/(n^3 +sqrt(n^(9)+1))

    sorry if it looks hard to read. I compared it to 1/n. And it diverged. The teachers said I must compare it to n^2/n^(9/2). NOw doing it this way makes the series in question converge!

  2. jcsd
  3. Dec 8, 2004 #2


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    Remember the LCT says that if you have two series a and b, and lim(n->infinity) a(n)/b(n)=s where s>0 and finite, then either "both a and b converge" or "both a and b diverge".

    What did you get for s when you did this limit and compared to 1/n? I'm guessing you got 0 or infinity (depending on which series is on top and bottom in the ratio)? This is what I get looking at your series and 1/n. The LCT does not apply when s is 0 or infinity... so we can't conclude that since 1/n diverges the given series diverges also.

    The trick is to pick a series so that when you take the ratio and then the limit, you get a finite value>0.

    Its been a while since I've done this stuff. Hope this helps. :smile:
  4. Dec 8, 2004 #3


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    learningphysics is completely correct.

    A little more detail: It is clear that, for large n, [itex]\frac{n^2}{n^3+\sqrt{n^9+1}}< \frac{1}{n}[/itex].

    Unfortunately since [itex]\Sigma \frac{1}{x}[/itex] diverges that tells you nothing about whether [itex]\Sigma\frac{n^2}{n^3+\sqrt{n^9+1}}[/itex] converges or diverges. You would need to show that your terms are larger than a series that diverges to know that it diverges or smaller than a series that converges to know that it converges.
  5. Dec 8, 2004 #4
    I get 1/2!!!!!! I will show my work! I still believe that they both diverge.

    Here it goes: Let an = n^2/(n^3 + sqrt(n^(9)+1); Let bn= 1/n.

    Now: (step 1) | n^2 * n |

    lim n->infinity | n^3 + sqrt(n^(9) +1) * 1 |

    (step 2) | n^3*(1) |

    | n^3 + sqrt(n^9(1+1/n^9))|

    lim n->infinity

    (step 3) |n^3*(1)|

    | n^3(1+sqrt(1+1/n^9))|

    lim n-> infinity

    (step 4)

    | 1|
    ___ = 1/2!!!!!!!!!!!!! AHHHHH!!!
    | 1+ sqrt(1)|

    <ps. sorry the divide signs would not show>
    Last edited: Dec 8, 2004
  6. Dec 9, 2004 #5
    Can anyone explain this contradiction? I would like to believe that their are no contradicitons in the realm of mathematics. Yet, I have not seem one statement that requires one to pick the test function be the highest powers found from the series in question. With that said, I wanted my test function to be 1/n.

    Any ideas?
  7. Dec 10, 2004 #6

    In step 3 I beleive it was, you took sqrt(n^9) to be n^3. This is wrong. You don't take the squareroot of the 9, you halve it, so the limit comparing to 1/n is 0, and the limit comparison test does not work for 1/n.

    Since the greatest power in the numerator is 2, and the greatest in the denominator is 9/2, you must compare to 1/n^(5.2), as your teacher said, in which case you will find it to converge.
  8. Dec 11, 2004 #7
    Thank you Thank you Thank you!!!! You are the best!
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