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idku

  • one year ago

I want to clarify something about the convergence of series and sequence.

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  1. idku
    • one year ago
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    So, for sequence to converge 1) Bound below and above 2) Monotonic (can be always decreasing or always increasing)

  2. idku
    • one year ago
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    and for series: 3) The limit [n→ infinity] A(n) = 0 (am I correct)

  3. zzr0ck3r
    • one year ago
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    yes, but the converse does not hold on 3)

  4. idku
    • one year ago
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    yes, what other conditions are there?

  5. idku
    • one year ago
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    well, yes not on those 2 - harmonic series.

  6. idku
    • one year ago
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    not those 3*

  7. idku
    • one year ago
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    But, what else must be true?

  8. zzr0ck3r
    • one year ago
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    of series?

  9. idku
    • one year ago
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    yes convergence of series

  10. zzr0ck3r
    • one year ago
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    there are many many theorems

  11. idku
    • one year ago
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    Oh, like Σ 1/n^p then p>1 geometric series |r|<1 (and that is where ratio test is coming from) and others....

  12. idku
    • one year ago
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    incl alternating series test.... and others

  13. idku
    • one year ago
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    so it basically depends on a series in every unique cse?

  14. idku
    • one year ago
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    case*

  15. zzr0ck3r
    • one year ago
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    pretty much. there are some nice ways for some families of functions, but in general there is nothing that will deal with them all.

  16. ganeshie8
    • one year ago
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    if you're comfortable with sequences, then you may think of any series as a sequence of partial sums

  17. zzr0ck3r
    • one year ago
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    I also will assume we are talking about the real numbers with the regular topology.

  18. idku
    • one year ago
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    yes, that is what i am reading on wiki and other sites ganeshie

  19. idku
    • one year ago
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    yes. no imaginary sequences. I am not considering anything crazy

  20. zzr0ck3r
    • one year ago
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    Getting good with sequences is a great idea if you are planning on doing analysis. There are definitions of continuity that uses sequences that make things much nicer than dealing with the standard \(\epsilon -\delta\) definition

  21. idku
    • one year ago
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    alright

  22. idku
    • one year ago
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    I was reading wiki just 3-5 minutes ago, and I saw that \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{(-1)^{n+1}}{n}=\ln(2)}\) (they don't post any link to an explanation)

  23. idku
    • one year ago
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    (i can create a new Q. if that is how the policy would play this time)

  24. idku
    • one year ago
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    (sorry if i am typing to much, don't be afraid to tell me to shut up:))

  25. zzr0ck3r
    • one year ago
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    This does not easily fall without other identities.

  26. idku
    • one year ago
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    which identities?

  27. zzr0ck3r
    • one year ago
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    series expansion of ln(x)

  28. ganeshie8
    • one year ago
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    criterian for convergence of an alternating series is an interesting one i remember seeing a very nice proof from analysis

  29. zzr0ck3r
    • one year ago
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    Maybe I am wrong, but I don't any basic ways to show this without more tools.

  30. zzr0ck3r
    • one year ago
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    You can find the taylor expansion and plug in two...

  31. zzr0ck3r
    • one year ago
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    But you are not proving anything....

  32. idku
    • one year ago
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    wait can i start from taylor series of 1/x and integrate?

  33. zzr0ck3r
    • one year ago
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    yeah

  34. zzr0ck3r
    • one year ago
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    but you already know you can do that, so what is the point :)

  35. zzr0ck3r
    • one year ago
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    unless you just feel like symbol pushing :)

  36. ganeshie8
    • one year ago
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    \[\frac 1{1+x}=\sum_{n=0}^\infty (-x)^n\] integrate both sides and plugin \(x=1\) this is fun but yeah doesn't give you the bigger picture of beatiful alternating series

  37. idku
    • one year ago
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    oh, we can get a series represenation for ln(x+1) :) nice

  38. idku
    • one year ago
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    \(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\)

  39. ganeshie8
    • one year ago
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    ln(x) has no hope at x=0 but surely it behaves well for all finite positive values of x

  40. idku
    • one year ago
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    oh yeah my index should have been n=0, i made that err

  41. idku
    • one year ago
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    and nice to hear it works for other ln values

  42. idku
    • one year ago
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    \(\large\color{black}{ \displaystyle \ln(x+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^nx^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=1 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}\)

  43. idku
    • one year ago
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    oh, sorry n=0

  44. idku
    • one year ago
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    \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\)

  45. ganeshie8
    • one year ago
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    now we're getting into radius of convergence and stuff for what values of x does that series converge ?

  46. idku
    • one year ago
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    ok, let me do ratio test:)

  47. idku
    • one year ago
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    |dw:1440308882162:dw|

  48. idku
    • one year ago
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    oh, |x|<1

  49. idku
    • one year ago
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    i left (-1)^n on purpose and hope correctly

  50. ganeshie8
    • one year ago
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    right, so its kinda tricky, the domain is restricted to [-1, 1)

  51. idku
    • one year ago
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    I will dare to sya in a face of god you are wrong (apologize in advance) I was just about say \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(1+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n1^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\) and because i know alternative harmonic converges, thus r=1 is valid in this case

  52. ganeshie8
    • one year ago
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    this is one of the most celebrated series, there are like more than 100 different proofs for its convergence... let me dig them up... :)

  53. idku
    • one year ago
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    we can make r=1, and just because if x=1 we get a result of "1" from the ratio test it doesn't tell us that series diverges, it is inconclusive when limit comes out with r=1, the domain is including 1.

  54. idku
    • one year ago
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    |dw:1440309328209:dw|this is kinda (a quite bad model) behavior of the partial sums of the alternating harmonic series

  55. ganeshie8
    • one year ago
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    do you mean : r = \(\color{Red}{-1}\) is in the interval of convergence because we know that alternating harmonic series converges

  56. idku
    • one year ago
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    oh, x=1, not r=1 i mean

  57. idku
    • one year ago
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    all i was trying to say x=1 is valid, and i guess i typed much more unneccessary stuff on top

  58. ganeshie8
    • one year ago
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    ahh okay you have that -1 in the front..

  59. idku
    • one year ago
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    for interval of convergence?

  60. ganeshie8
    • one year ago
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    \(\sum\limits_{n=0}^{\infty}\dfrac{x^n}{n}\) converges for all \(x\in [-1, 1)\)

  61. idku
    • one year ago
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    yes, but when it is multiplied times (-1)^n then the interval is x= [-1,1] , isn't it?

  62. ganeshie8
    • one year ago
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    right, we're on same page :)

  63. idku
    • one year ago
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    oh, and then we extract that representation \(\large\color{black}{ \displaystyle \ln(\color{red}{1}+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n\color{red}{1}^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\sum_{ n=1 }^{ \infty } \frac{(-1)^{n+1}}{n}}\)

  64. idku
    • one year ago
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    now that series makes sense ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 ...

  65. idku
    • one year ago
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    thank you:)

  66. idku
    • one year ago
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    well, should be (1)^(n\(\color{red}{-}\)1), but it is all sa,e

  67. ganeshie8
    • one year ago
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    looks good! its not a big deal, but below looks better : \(\large\color{black}{ \displaystyle \ln(\color{red}{x}+1)=\sum_{ n=0 }^{ \infty } \frac{(-1)^n\color{red}{x}^{n+1}}{n+1}}\) \(\large\color{black}{ \displaystyle \lim\limits_{x\to 1}~\ln(\color{red}{x}+1)=\lim\limits_{x\to 1}~\sum_{ n=0 }^{ \infty } \frac{(-1)^n}{n+1}}\) \(\large\color{black}{ \displaystyle \ln(2)=\lim\limits_{x\to 1}~\sum_{ n=1 }^{ \infty } \frac{(-1)^{n}x^n}{n}}\)

  68. idku
    • one year ago
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    why is that better if I am permitted to plug in x=1 any way?

  69. ganeshie8
    • one year ago
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    because when deriving that series representation, before integrating, we have used geometric series restricting ourselves to \((-1,1)\)

  70. ganeshie8
    • one year ago
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    so technically we're not allowed to "plugin" x=1

  71. idku
    • one year ago
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    oh, ok.... but we can always trick the math rules:) by the way you last line shouldn't have x^n on top, correct?

  72. idku
    • one year ago
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    no nvm

  73. idku
    • one year ago
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    well it is still same you are plugging in x=1, but you are technically not:O

  74. ganeshie8
    • one year ago
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    limit of an infinite sum is not same as sum of limit

  75. ganeshie8
    • one year ago
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    again, technically, we're not allowed to swap limit and infinite sum

  76. ganeshie8
    • one year ago
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    thats the reason i left it as \(x\to 1\)

  77. idku
    • one year ago
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    but plugging x=1 does not make the series diverge tho' the ratio test is 1 is x=1 (and jsut that no conclusions can be made, but not that it will make the sum diverge). THis series is conditionally convergent when x=1. I don't really see much problem with actually going ahead and plugging x=1. It is just illogical to me-:( :)

  78. idku
    • one year ago
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    the ratio test is 1, *if* x=1

  79. idku
    • one year ago
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    maybe i can re-express myself

  80. ganeshie8
    • one year ago
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    right, right, the variables are different so we can take the limit, no issues i guess

  81. idku
    • one year ago
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    alrighty:) I guess then I am done for now

  82. idku
    • one year ago
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    Will use a correct notation lim [n→∞] (thanks)^n

  83. ganeshie8
    • one year ago
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    lookup the proof for convergence of alternating series, its really neat... im sure you will enjoy all other proofs of various convergence tests..

  84. idku
    • one year ago
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    ok, lets see what they got;)

  85. ganeshie8
    • one year ago
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    here it is https://www.youtube.com/watch?v=tt430qiyIds&index=19&list=PL04BA7A9EB907EDAF just watch it like a movie, don't wry about all the weird terms that you might hear... :)

  86. idku
    • one year ago
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    oh:) It's 2:30 am in my location, i was sitting late at night because I was curious... this movie is too long for now. i have to go offline right now:) I saved the link (and if anything i can login to see the link)..... thank you very much! C(U)

  87. idku
    • one year ago
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    gtg, sry bye

  88. ganeshie8
    • one year ago
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    have good sleep :)

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