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unicwaan

  • one year ago

Determine whether the sequence converges or diverges. If it converges, give the limit. 60, -10, 5/3, -5/18, ...

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  1. SolomonZelman
    • one year ago
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    ok, your common ratio is>?

  2. unicwaan
    • one year ago
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    Okay

  3. SolomonZelman
    • one year ago
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    your common ratio is not "okay" lol

  4. SolomonZelman
    • one year ago
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    do you know what a common ratio is?

  5. unicwaan
    • one year ago
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    XD Oh I thought you were telling me that, I'm sorry. I do not know what a common ratio is.

  6. SolomonZelman
    • one year ago
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    common ratio, is a number by which you multiply to get to the next term each time.

  7. unicwaan
    • one year ago
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    Oh okay so would it be (-1/6)?

  8. SolomonZelman
    • one year ago
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    So, if I had 1, 2, 4, 8, 16 .... then my common ratio is 2 (and common ratio is denoted as r, so in this example r=2) --------------------------------------------- Yes, r=-1/6 is right

  9. unicwaan
    • one year ago
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    Okay! I undertsand that!

  10. SolomonZelman
    • one year ago
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    Note: This is a geometric series (because it is multiplied times some number (referred to as common ratio (r) ).

  11. SolomonZelman
    • one year ago
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    Do you know what it means for a series to converge and for a series to diverge?

  12. SolomonZelman
    • one year ago
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    oh, for a sequence to diverge and converge

  13. unicwaan
    • one year ago
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    No I do not, that's what I'm coversed on. My lesson never went over it :(

  14. unicwaan
    • one year ago
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    confused**

  15. SolomonZelman
    • one year ago
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    sequence, is just a list of terms. When you say that "sequence converges to 3" (for example 3) you are saying that as you take 1000th 100000th (and roughly speaking infinity-th terms) they all will be approximately 3 (and each time they will get closer and closer to 3)

  16. SolomonZelman
    • one year ago
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    Say you started from 60, and you were to just divide by 1/6 (not -1/6). \(a_1=60\) \(a_2=10\) \(a_3=5/3\) \(a_4=5/18\) \(a_5=5/108\) and roughly speaking \(a_\infty=5/\infty=0\)

  17. unicwaan
    • one year ago
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    Ohhhhhh that makes sense okay

  18. SolomonZelman
    • one year ago
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    No, you won't actually hit 0. Never! But you closer and closer approach to zero every time when you divide by 1/6. (this is the idea of a "limit" that you approach some value)

  19. unicwaan
    • one year ago
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    What about if something diverges?

  20. SolomonZelman
    • one year ago
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    now, in our case we are multiplying (in the first post, should say multiplying too) by -1/6 so, all we change is that \(a_1=60\) \(a_2=-10\) \(a_3=5/3\) \(a_4=-5/18\) \(a_5=5/108\) and so on.... we still get \(a_\infty=\pm5/\infty=0\) (whether a positive or negative number is divided by infinity, you get 0 in either case)

  21. SolomonZelman
    • one year ago
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    So the more terms you take, the close and close you approach what value?

  22. unicwaan
    • one year ago
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    0, since the number values get smaller and smaller

  23. SolomonZelman
    • one year ago
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    yes, that is correct

  24. SolomonZelman
    • one year ago
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    So that means that "sequence converges to zero"

  25. unicwaan
    • one year ago
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    ^-^ Thanks a lot that makes sense!

  26. SolomonZelman
    • one year ago
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    (and that is the only case - i.e. - sequence converges to zero - for a series to converge. If you want we can talk about the convergence of a series as well (later on in this thread))

  27. SolomonZelman
    • one year ago
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    And you asked about divergence, i will answer that now

  28. unicwaan
    • one year ago
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    When I do reach series in my lesson I will come to you! and yes

  29. SolomonZelman
    • one year ago
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    Say you got 1, 2, 4, 8, 16, 32 ....... what is the pattern here, can you tell me?

  30. unicwaan
    • one year ago
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    the next number is being multiplied by 2

  31. SolomonZelman
    • one year ago
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    yes, r=2

  32. SolomonZelman
    • one year ago
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    And as you take more and more terms you multiply times 2 more and more.... so you are going to go into infinity. (makes sense?)

  33. unicwaan
    • one year ago
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    Yes!

  34. SolomonZelman
    • one year ago
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    that means that sequence diverges.

  35. unicwaan
    • one year ago
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    Oh okay so diverges means the increase (or decrease) infinitely while converges ia to get closer and closer to a certain number?

  36. SolomonZelman
    • one year ago
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    so sequence can diverge in 2 cases. (1) if you have an alternating sequence that is not approaching zero. Such that \(A_n=4(-1)^n\), so your terms are going to be 4, -4, 4, -4, and so on.... and as n approaches infinity (n→∞) you don't know what your term is going to be because it can be either 4 or -4. (2) if your sequence goes into infinity (or negative infinity).

  37. SolomonZelman
    • one year ago
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    yes, what you said is correct

  38. SolomonZelman
    • one year ago
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    that is a good definition of convergence of a SEQUENCE.

  39. unicwaan
    • one year ago
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    Okay only for a sequence, got it.

  40. SolomonZelman
    • one year ago
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    now, a series is the sum of terms in a sequence.

  41. SolomonZelman
    • one year ago
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    think about it. Series is a sum of all terms in the sequence. So if sequence converges to 3 to -8 (or to anything besides 0) then you are going to be adding 3's -4's or something's forever and that will not have a defined sum (will go into ±∞)

  42. SolomonZelman
    • one year ago
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    is this making some sense?

  43. unicwaan
    • one year ago
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    I believe so, yes

  44. SolomonZelman
    • one year ago
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    yes, so for example in your case, your sequence converges to what?

  45. unicwaan
    • one year ago
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    The sequence converges to zero

  46. SolomonZelman
    • one year ago
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    yes

  47. SolomonZelman
    • one year ago
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    and the rule is that a geometric SERIES (a series that has all terms follow the pattern of multipling times some common ratio) will always converge if common ratio r is: -1>r>1

  48. unicwaan
    • one year ago
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    But in this case, the rule does not apply

  49. SolomonZelman
    • one year ago
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    this rule always applies provided that ratio is: -1<r<1

  50. SolomonZelman
    • one year ago
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    and so it is in your case, cuz r=-1/6 so it is between -1 and 1

  51. unicwaan
    • one year ago
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    Oh okay, I was thinking 0 was r, but r is the common ratio which was -1/6, That makes sense

  52. SolomonZelman
    • one year ago
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    lets see, but why does the common ratio r has to satisfy -1<r<1? I will adress that now....

  53. SolomonZelman
    • one year ago
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    for any geometric sequence you multiply times r to get to the next term. So lets say you start from \(a_1=5\) (or from any non-zero first term) \(\large\color{black}{ \displaystyle a_1=5}\) \(\large\color{black}{ \displaystyle a_2=5\cdot r}\) \(\large\color{black}{ \displaystyle a_3=5\cdot r^2}\) \(\large\color{black}{ \displaystyle a_4=5\cdot r^3}\) so on.... \(\large\color{black}{ \displaystyle a_n=5\cdot r^{n-1} }\)

  54. SolomonZelman
    • one year ago
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    and then you get, roughly speaking \(\large\color{black}{ \displaystyle a_\infty=5\cdot r^{\infty -1}}\)

  55. unicwaan
    • one year ago
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    Makes sense

  56. SolomonZelman
    • one year ago
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    so, if r\(\ge\)1 or if r\(\le\)-1 then the \(\large\color{black}{ \displaystyle r^{\infty -1}}\) part will go to either ∞ or -∞. but if -1<r<1 then you are going to have \(\large\color{black}{ \displaystyle r^{\infty -1}=0}\)

  57. SolomonZelman
    • one year ago
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    your teacher will go over in detail, but the bottom line is that series is convergent, ONLY when sequence converges to 0. (there are many more tests for various series to converge/diverge that you will learn later in calc, but for geometric series, all you need to know is sequence converges to 0) and sequence convergence is when the \(a_\infty\) approaches one particular value.

  58. SolomonZelman
    • one year ago
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    And for any geometric series \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ A_n}\) the sum is given by \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ A_n~=\frac{a_1}{1-{\rm r}}}\) in your case, the terms 60 is \(a_1\) and r=-1/6 so, \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ A_n~=\frac{60}{1-{\rm -\dfrac{1}{6}}}=\frac{60}{\dfrac{7}{6}}=360/7 }\)

  59. unicwaan
    • one year ago
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    :D Thank you for explaining so much to me! This really helped a lot. and I learn about that and summantion notation in my next lesson so thanks!

  60. SolomonZelman
    • one year ago
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    Anytime!

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