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Why does it matter what convergence test I use?

  1. Apr 11, 2016 #1
    I just took a calc 2 test and got 3/8 points on several problems that asked you to show convergence or divergence. The reason being that I didn't use the correct test of convergence? The answer was right, if you get to the point where you know the series converges, then why does it matter which test I used?

    Does anyone have any tips for how to tell which test I should be using when I look at a sequence or series? (my idea is- when in doubt, use ratio test).
     
  2. jcsd
  3. Apr 11, 2016 #2

    SammyS

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    You may have found the correct value for what the series converged to, without actually showing that the series converged.

    Can you give examples of the questions asked?
     
  4. Apr 11, 2016 #3
    I can see two possibilities.
    1. Your answers were not actually correct and/or your method did not apply to the problem.
    2. The test asked you to solve a problem in a specific way and you didn't.

    It would help to have an example of why you lost points.
     
  5. Apr 11, 2016 #4

    micromass

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    Getting the correct answer is not so difficult. You can basically guess "convergent" and have a 1 in 2 chance of being correct. What matters is that your reasoning is correct. If somebody said the correct answer but with incorrect reasoning, I wouldn't give any points.

    So, are you sure your reasoning is correct?
     
  6. Apr 11, 2016 #5
    The answer was correct (converges), but the method by which I got there was not.
    2vvuxxg.jpg
     
  7. Apr 11, 2016 #6

    micromass

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    The series diverges actually.
     
  8. Apr 11, 2016 #7
    Haha.... how so? My instructor said answer was right, method was not. Very interesting... :s
     
  9. Apr 11, 2016 #8
    Here is the question in full, parts a, b, and e (above). 3 points for the "correct answer" and minus 5 for incorrect method as indicated by my instructor.

    1zokw79.jpg
     
  10. Apr 11, 2016 #9

    SammyS

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    ##\displaystyle \frac{3k^2+2}{k^3+2k+5} \ ## behaves like ##\displaystyle\ \frac1k \ ## for large ##k## .
     
  11. Apr 12, 2016 #10
    I posted two photos below showing how to determine convergence or divergence.
    The first photo proves why the divergence test is inconclusive for the series you provided.
    The second photo shows how to prove divergence of the series you provided by means of the integral test.
    Hope this helps. Let me know if the photos are not displaying please, and I will do what I can to fix it.
    https://scontent.fsnc1-1.fna.fbcdn.net/v/t1.0-9/12994327_10209359465651426_1265712068729596108_n.jpg?oh=e24d119a922a506c62295f24cab018b6&oe=577ED737
    https://scontent.fsnc1-1.fna.fbcdn.net/hphotos-xta1/v/t1.0-9/12439374_10209359465611425_8823074394851131369_n.jpg?oh=f81f8f11dddc2333cb813384a1f93226&oe=577CEFCA
     
  12. Apr 12, 2016 #11

    Samy_A

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    For a), the question was about the sequence, not the series.

    For b), your answer is not very readable.
    But you don't mention the test you are using (hence -5). And the general term of the series behaves like ##2k^{-\frac{5}{3}}## for large k.
     
  13. Apr 12, 2016 #12

    pwsnafu

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    At my university, we obtain very detailed marking schemes. Here's what we would require from a Calc 2 student (sample answer in black, marking scheme in blue).
    Granted, different schools have different standards, but this should give you some expectations.
    Notice in part (a) its asking for the sequence not series.
    (a)

    Observe ##\frac{-5n^2}{3n^3-2n+4} \leq \frac{(-1)^n 5n^2}{3n^3-2n+4} \leq \frac{5n^2}{3n^2-2n+4}##.
    1 method mark for using squeeze theorem.
    1 accuracy mark for correct upper and lower bounds.

    Then
    ##\lim_{n\to\infty} \frac{5n^2}{3n^3 - 2n +4} = \lim_{n\to\infty} \frac{5/n}{3-2/n^2+4/n^3}##
    ##=\frac{\lim_{n\to\infty} 5/n}{\lim_{n\to\infty}(3-2/n^2+4/n^3)}## by limit laws
    ##=\frac{0}{3-0+0}## by standard limit ##1/n \to 0##
    ##=0.##
    1 method mark for dividing by highest power.
    1 justification mark for stating "limit laws" and "standard limits"

    1 accuracy mark for obtaining limit equal to 0.
    Similarly, ##\frac{-5n^2}{3n^3-2n+4} \to 0##
    So by squeeze theorem, ##\lim_{n\to\infty} \frac{(-1)^n 5n^2}{3n^2-2n+4} = 0##.
    Hence because 0 is a real number, the sequence converges.
    1 justification mark for stating squeeze theorem.
    1 accuracy mark for concluding that sequence converges.
    1 notation mark (must be 100% correct).
    Total 8 marks.


    (b)
    Let ##a_k = \frac{2k+3}{\sqrt[3]{k^8+9k}}##. Then
    ##\sum_{k=1}^\infty a_k < \sum_{k=1}^\infty \frac{2k+3k}{\sqrt[3]{k^8+0}} = \sum_{k=1}^\infty \frac{5}{k^{5/3}}##.
    1 method mark for using comparison test.
    1 accuracy mark for finding correct upper bound.

    Now ##\sum_{k=1}^\infty \frac{1}{k^{5/3}}## converges as it is a harmonic p-series with ##p=5/3 > 1##.
    1 justification mark for mentioning harmonic p-series.
    Hence by comparison test, ##\sum_{k=1}^\infty a_k## also converges.
    1 accuracy mark for stating the series converges.
    1 justification mark for mentioning comparison test.

    As ##a_k > 0## for all ##k##, ##\sum_{k=1}^\infty a_k## converges absolutely.
    1 accuracy mark for stating the series converging absolutely with suitable reason.
    1 notation mark (must be 100% correct).
    Total 7 marks.


    (e)
    Let ##a_k = \frac{3k^2+2}{k^3 + 2k +5}##. Then
    ##\sum_{k=1}^\infty a_k > \sum_{k=1}^\infty\frac{3k^2}{k^3+2k^3+5k^3} = \frac{3}{8}\sum_{k=1}^\infty\frac{1}{k}##.
    1 method mark for using comparison test.
    1 accuracy mark for finding correct upper bound.

    Then ##\sum_{k=1}^\infty\frac{1}{k}## diverges because it is the harmonic p-series with ##p=1##.
    1 justification mark for mentioning harmonic p-series diverges.
    Hence, by the comparison test, ##\sum_{k=1}^\infty a_k## diverges.
    1 accuracy mark for stating the series diverges.
    1 justification mark for stating comparison test.
    1 notation mark (must be 100% correct).
    Total 5 marks.
     
  14. Apr 12, 2016 #13
    I first taught myself to recognize and name series (geometric series, telescoping series, harmonic series, p-series, alternating series, etc). All of these series have their own set of conditions and rules to follow, so it is difficult to identify which series you're looking at at first (but keep with it and study the theorems of each series and you'll quickly be able to recognize which type of series you are looking at).
    I then practiced by working out multiple exercises that had varying series.
    I asked multiple questions.
    Lastly, I used note cards to define each type of series and I wrote out a step-by-step method to determine the convergence (or divergence) of each series.

    If I couldn't tell which type of series I was looking at I followed the following advice (which I read while doing my home work on Pearson's MyMathLab):
    "If the general kth term of the series involves k!, k^k, or a^k, where a is a constant, the Ratio Test is advisable. Series with k in an exponent may yield to the Root Test. If the general kth term of the series is a rational function of k (or a root of a rational function), use the Comparison Test or the Limit Comparison Test."

    Lastly, to answer your question (which is the title of your thread), the convergence test you use matters because it either proves, or disproves, what our intuition tells us about a given series. Because each type of series has its own conditions and rules, we have to be careful which convergence test we use. For instance, its easy to think that the harmonic series converges because the limit of its underlying sequence goes to zero. However, despite some people's intuition, we can use the p-series rules to prove that the harmonic series diverges. Furthermore, we can support our conclusions from the p-series by performing the integral test on the harmonic series. In reality, although the limit of the underlying sequence goes to zero (and looks like it converges to zero), it doesn't go to zero fast enough and therefore diverges. In short, we can use convergence tests to support or refute our intuition (because lets face it, we're all human and our intuition is not always right).
     
    Last edited: Apr 12, 2016
  15. Apr 12, 2016 #14
    "The answer was right, if you get to the point where you know the series converges, then why does it matter which test I used?"

    This is a very basic question that all students would benefit from knowing the answer to.

    You may know the right answer, but how does the grader know that you actually knew the right answer instead of just guessing, or perhaps that you may have used incorrect reasoning to arrive at the answer? Especially in a case like convergence / divergence where there are only two answers, anyway?

    That's why we profs want math students to show their work. This also enables us to help you get on the right track if we know what kind of mistake you made.
     
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