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Identifying Types of Converging Series

  1. Aug 3, 2015 #1
    Hello, I'm looking for some help with this problem for my Calculus 2 class. Since it's a summer class my professor wasn't able to explain everything fully so if you can help me that would be great : )

    1. The problem statement, all variables and given/known data

    Select the FIRST correct reason why the given series converges.

    A. Convergent geometric series
    B. Convergent p series
    C. Comparison (or Limit Comparison) with a geometric or p series
    D. Alternating Series Test
    E. None of the above

    1. Σn=1 6(4)n/72n
    2. Σn=1 sin2(6n)/n2
    3. Σn=1 cos(nπ)/ln(2n)
    4. Σn=1 (−1)n/(6n+5)
    5. Σn=1 (n+1)(24)n/52n
    6. Σn=1 (−1)n * √(n)/(n+9)

    2. Relevant equations
    Geometric Series:
    Σn=1 arn-1 will converge if -1 < r < 1

    P Series:
    Σn=1 1/np converges if p > 1

    Alternating Series Test:
    1) bn+1 </= bn for all n and
    2) limn->∞ bn = 0

    3. The attempt at a solution
    1. A
    2. B
    3. E
    4. D
    5. C
    6. D
     
    Last edited by a moderator: Aug 3, 2015
  2. jcsd
  3. Aug 3, 2015 #2

    Ray Vickson

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    I get (in order 1-6) A, C, D, D, E, D.
     
  4. Aug 4, 2015 #3
    My professor told me that I got numbers 2 and 3 correct
     
  5. Aug 4, 2015 #4

    Ray Vickson

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    I disagree. Here are the details.
    1. ##\sum 6 \;4^n/7^{2n} = 6 \sum (4/49)^n##, so geometric (A).
    2. In ##\sum \sin^2 (6n) /n^2## the terms are (i) not geometric; (ii) not alternating; (iii) not in the simple form ##1/n^p##. However, ##0 \leq \sin^2(6n)/n^2 \leq 1/n^2##, so a comparison with the series ##1/n^2## works (C).
    3. ##\sum \cos(n \pi)/ \ln(n) ## is an alternating series, since ##\cos(n \pi) = (-1)^n##. So, D.
    4. ##\sum (-1)^n / (6n+5)## is alternating, so D.
    5. ##\sum (n+1) \; 24^n / 5^{2n} = \sum (n+1) r^n, \;\; r = 24/25##. Here, A,B,C,D do not apply, and that leaves E.
    6. ##\sum (-1)^n \sqrt{n}/(n+9)## (or ## \sum (-1)^n \sqrt{ n/(n+9)}##---can't tell which you mean) is alternating in either interpretation, so D.
     
  6. Aug 4, 2015 #5
    The correct answer ended up being:

    1. A
    2. C
    3. D
    4. D
    5. C
    6. D

    Thank you for your help : )
     
  7. Aug 4, 2015 #6

    Ray Vickson

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    How do you get 5 C? Did you read my detailed explanation?
     
  8. Aug 4, 2015 #7

    micromass

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    Why wouldn't the comparison test apply in ##5##? You can do ##n+1\leq \alpha^n## for some small enough ##\alpha>1##.
     
  9. Aug 4, 2015 #8

    Ray Vickson

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    OK. That would be an answer to my question (why?) that the OP did not answer. For all I know the OP may have used false reasoning.
     
  10. Aug 4, 2015 #9
    I don't want to assume, but that sounded a bit rude.

    Still, your explanations are helpful.

    I am assuming that going off of your explanation for 5 that we can compare that to a geometric function ∑n=0 anrn and since the absolute value of r is less than 1 we can go ahead and reason that the limit would be equivalent to a/(1-r)
     
  11. Aug 5, 2015 #10

    Ray Vickson

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    No: a geometric series would be ##\sum r^n## (or ##\sum c r^n = c \sum r^n## for constant ##c##), but ##\sum a_n r^n## is NOT a geometric series if ##a_n## varies with ##n##. However, if ##a_n \to c = \text{constant} ## as ##n \to \infty##, then comparison with the series ##\sum c r^n## can be made to work without too much effort (although some extra effort is needed), However, in your case you have ##a_n = n+1 \to +\infty## as ##n \to \infty##, so that type of argument does not work automatically. If you thought it did, that is what what I would call "false reasoning".

    However, as 'micromass' has pointed out, if you choose a small enough ##\epsilon > 0## so that ##(1+\epsilon) r < 1##, then for some finite ##N> 0## we have ##n+1 \leq (1+\epsilon)^n## for all ##n \geq N##, so for ##n \geq N## we have ##0 \leq (n+1) r^n \leq [(1+\epsilon) r]^n## and we can compare with the geometric series ##\sum [(1+\epsilon) r]^n##. Note, however, that some such argument as this must be acknowledged and recognized in order for C to be a valid and well-founded answer. Without this realization, just saying it would, again, be false reasoning.
     
    Last edited: Aug 5, 2015
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