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blackstreet23

  • one year ago

Find the radius and the interval of convergence sigma (X^k)/(k+1)

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  1. amistre64
    • one year ago
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    first take the ratio, and work the limit

  2. blackstreet23
    • one year ago
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    \[\sum_{k=0}^{\infty} \frac{ X^k }{ k+1 }\]

  3. blackstreet23
    • one year ago
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    I did that well. I am having troubles testing the limits

  4. amistre64
    • one year ago
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    hmm, show your work, lets see how well you got to that point then

  5. blackstreet23
    • one year ago
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    I dont understand how to test if it decreasing in the alternating series test

  6. amistre64
    • one year ago
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    so your trying to test for x=-1 ?

  7. blackstreet23
    • one year ago
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    give me a second

  8. amistre64
    • one year ago
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    \[\lim_{k\to\infty}\frac{x^{n+1}/(k+2)}{x^n/(k+1)}\] \[\lim_{k\to\infty}x\frac{k+1}{k+2}\] \[|x|\lim_{k\to\infty}\frac{k+1}{k+2}=|x|\]

  9. blackstreet23
    • one year ago
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  10. amistre64
    • one year ago
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    when x=1, this is a harmonic series ... when x=-1, its an alternating harmonic series ... do you recall the convergence of those?

  11. blackstreet23
    • one year ago
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    harmonic no

  12. blackstreet23
    • one year ago
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    how do you do it?

  13. amistre64
    • one year ago
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    the harmonic does not converge, the alternating does ...

  14. blackstreet23
    • one year ago
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    but why? Could you explain?

  15. blackstreet23
    • one year ago
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    what is a harmonic series? and how do i know if it converges or diverges?

  16. amistre64
    • one year ago
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    harmonic is\[\frac11+\frac12+\frac13+\frac14+\frac15+...\] the proofing is textbook, but it eludes me

  17. blackstreet23
    • one year ago
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    so is like \[\frac{ 1 }{ K! }\]

  18. amistre64
    • one year ago
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    http://web.williams.edu/Mathematics/lg5/harmonic.pdf

  19. blackstreet23
    • one year ago
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    ohh my bad \[\frac{ 1 }{ k }\]

  20. blackstreet23
    • one year ago
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    ohh ok i see if it has a K on the bottom

  21. beginnersmind
    • one year ago
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    @amistre "the proofing is textbook, but it eludes me" 1/3 + 1/4 > 1/2 1/5 + 1/6 + 1/7 + 1/8 > 1/2 1/9 + ... 1/16 > 1/2 etc.

  22. blackstreet23
    • one year ago
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    I still dont get how can someone recognize that a function is a harmonic series just by looking at it

  23. amistre64
    • one year ago
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    the alternating harmonic series converges :) that proof eludes me as well ... just recall it from the classes.

  24. blackstreet23
    • one year ago
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    is because of the K?

  25. amistre64
    • one year ago
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    1/n is the harmonic series by definition

  26. amistre64
    • one year ago
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    yes

  27. blackstreet23
    • one year ago
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    so K by itself would be a harmonic series?

  28. blackstreet23
    • one year ago
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    2K +1

  29. amistre64
    • one year ago
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    1/k^p converges for p > 1

  30. blackstreet23
    • one year ago
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    10 k -2

  31. blackstreet23
    • one year ago
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    isnt it 1/k^p a p-series?

  32. amistre64
    • one year ago
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    yes

  33. blackstreet23
    • one year ago
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    ok sooo if it has a K is a harmonic series and if it has an exponent is a p series?

  34. amistre64
    • one year ago
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    1/k = 1/k^1 by definition, 1/k is the harmonic series.

  35. amistre64
    • one year ago
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    that is why the p series test says that if p > 1 it converges, but if 0<p<= 1 it diverges

  36. blackstreet23
    • one year ago
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    so it diverges?

  37. amistre64
    • one year ago
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    why do i feel like this is just going in circles ....

  38. blackstreet23
    • one year ago
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    and please special help in the alternating series one !

  39. blackstreet23
    • one year ago
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    sorry i am just reinforcing the knowledge you are giving me by enphasing what you say lol

  40. amistre64
    • one year ago
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    the alternating harmonic series converges, by properties that i cant recall either. they are explained better by the texts

  41. amistre64
    • one year ago
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    http://www.math.com/tables/expansion/tests.htm

  42. amistre64
    • one year ago
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    radius = 1 interval is: [-1,1)

  43. blackstreet23
    • one year ago
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    Ohh ok. I am so sleepy right now haha. I will analyse it more tomorrow :). Thanks a lot ! anything I will send you more messages :D

  44. amistre64
    • one year ago
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    im getting tired as well, but there are plenty of smarter people then me about :)

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