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anonymous

  • one year ago

I have a issure with a Series exercises,anyone can help please?

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  1. anonymous
    • one year ago
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    \[\sum_{1}^{\infty} \frac{ n ^{2} }{ 2^{n} }\]

  2. anonymous
    • one year ago
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    @ayeshaafzal221

  3. IrishBoy123
    • one year ago
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    are you trying to establish whether the series converges/diverges?

  4. IrishBoy123
    • one year ago
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    Does \[\lim\limits_{n→∞} a_n = 0\]

  5. anonymous
    • one year ago
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    I'm trying to reach the final result

  6. anonymous
    • one year ago
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    I wanna see how to do that

  7. IrishBoy123
    • one year ago
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    you are adding the terms in a series but if they are increasing as you go there is no way the series can converge to a finite sum so look at the individual term. it just looks bad but you can apply l'Hopital's rule:

  8. IrishBoy123
    • one year ago
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    sorry it looks good as polynomial on top, exponent on bottom, my bad do you know l'Hospital?

  9. IrishBoy123
    • one year ago
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    i am looking at this list, as you seem to need some technique or approach i'd suggest we just work down it http://www.math.hawaii.edu/~ralph/Classes/242/SeriesConvTests.pdf

  10. anonymous
    • one year ago
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    Yeah,the basics,I derivated each term and try to do the limit again

  11. anonymous
    • one year ago
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    derivate*

  12. IrishBoy123
    • one year ago
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    cool do it!

  13. anonymous
    • one year ago
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    But (n^2)' = 2n

  14. IrishBoy123
    • one year ago
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    yes do the bottom next

  15. anonymous
    • one year ago
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    and I dont know how to derivate 2^n

  16. Jhannybean
    • one year ago
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    \[\sf a^x = a^x \cdot \ln a\]

  17. Jhannybean
    • one year ago
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    oop... \(\frac{d}{dx}\)* of..... haha.

  18. anonymous
    • one year ago
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    Oh

  19. anonymous
    • one year ago
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    so it's (2n)/(2^n) * ln 2?

  20. anonymous
    • one year ago
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    and it is infinite / infinite and I do lhospital again?

  21. IrishBoy123
    • one year ago
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    yes

  22. anonymous
    • one year ago
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    0/(2^n)*ln 2 = 0

  23. anonymous
    • one year ago
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    this is the answer?

  24. IrishBoy123
    • one year ago
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    so what does that prove?

  25. anonymous
    • one year ago
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    It is convergent?

  26. IrishBoy123
    • one year ago
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    no!!!

  27. anonymous
    • one year ago
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    Dammit 50:50 chance and I failed

  28. IrishBoy123
    • one year ago
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    we applied the test to the individual term of the series and so the terms get smaller as this series goes on but it does not follow that the series is finite look at the link again http://www.math.hawaii.edu/~ralph/Classes/242/SeriesConvTests.pdf we have applied the first test. it can tell you something is not convergent but it cannot tell you whether something is convergent

  29. Jhannybean
    • one year ago
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    p-series?

  30. IrishBoy123
    • one year ago
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    An example: \[\Sigma \frac{1}{n} \] diverges even though the individual terms get smaller and smaller

  31. anonymous
    • one year ago
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    I see,because the "p" equals 1,right?

  32. Jhannybean
    • one year ago
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    Oh because according to the p-series test, \(\dfrac{1}{n^{\color{red}{1}}} \) means that 1 \(\gt\) 1, so it diverges.

  33. Jhannybean
    • one year ago
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    Ahh, i meant not greater than.

  34. Jhannybean
    • one year ago
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    There is no darn symbol for that.

  35. IrishBoy123
    • one year ago
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    @Jhannybean i'm literally blagging it from that sheet i linked, and it says Does \(a_n = 1/n^p, n ≥ 1?\), but we have Does \(a_n = n^2/n^p, n ≥ 1?\) i'd ratio test it as i feel safer

  36. IrishBoy123
    • one year ago
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    @xfire30 you there and OK ?!

  37. anonymous
    • one year ago
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    Yes,just to make sure,if the result was 0,it was divergent no matter the type of the series,but if it is 0,it depends on the type of the series,right?

  38. IrishBoy123
    • one year ago
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    if \[ \lim\limits_{n→∞} a_n \ne 0 \] it diverges

  39. IrishBoy123
    • one year ago
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    but if \[\lim\limits_{n→∞} a_n = 0\] on we plod, it could still go either way

  40. IrishBoy123
    • one year ago
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    so we need another test can you try this one from the sheet? RATIO TEST Is limn→∞ |an+1/an| 6= 1?

  41. anonymous
    • one year ago
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    I will try,But I have to go for 20 minutes to pick up my broather from the train station,I will tag you when I'm back

  42. IrishBoy123
    • one year ago
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    is \[ \lim\limits_{n→∞} |\frac{a_{n+1}}{a_n}| \lt 1\]??? i have to go too but try this out and see where you go

  43. anonymous
    • one year ago
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    No,it is bigger than 1

  44. IrishBoy123
    • one year ago
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    you sure? \[\large \frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}} = \frac{(n+1)^2}{2^{n+1}}.\frac{2^n}{n^2}\]

  45. anonymous
    • one year ago
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    I dont get it,is this another exercise or the same one?

  46. IrishBoy123
    • one year ago
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    this is the ratio test you were going to apply

  47. anonymous
    • one year ago
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    I dont get it

  48. anonymous
    • one year ago
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    How did you got that?

  49. IrishBoy123
    • one year ago
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    |dw:1442256369176:dw|

  50. IrishBoy123
    • one year ago
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    |dw:1442256441060:dw|

  51. IrishBoy123
    • one year ago
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    work that through to a conclusion :p

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