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SolomonZelman

  • one year ago

just made it up....

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  1. SolomonZelman
    • one year ago
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    List of weirdest questions that have seen....i am all making it up \(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n!)^n}{n^{(n!)}}}\) \(\Large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \rm \frac{(n^2+1)!}{n^n}}\)

  2. xapproachesinfinity
    • one year ago
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    eh lolz the first converge

  3. SolomonZelman
    • one year ago
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    should...

  4. mathmath333
    • one year ago
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    i find it normal though i dont undestand it completly.

  5. xapproachesinfinity
    • one year ago
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    eh you made up those series haha let see how the first can converge

  6. xapproachesinfinity
    • one year ago
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    let me first cheat with wolf

  7. mathmath333
    • one year ago
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    lol

  8. SolomonZelman
    • one year ago
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    \(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}\)

  9. SolomonZelman
    • one year ago
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    it is a very good latex practice too

  10. SolomonZelman
    • one year ago
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    + i haven't done them in a while... good start to refresh them

  11. xapproachesinfinity
    • one year ago
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    no i can't bear writing all that! too lazy

  12. SolomonZelman
    • one year ago
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    \(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\left|\frac{(n+1)!~^{n+1}}{(n+1)^{(n+1)!}}\times \frac{n^{(n!)}}{(n!)^n}\right|}\) \(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\) btw dit is ratio test

  13. xapproachesinfinity
    • one year ago
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    looks doable to me! wolf kicked me out lol didn't give me an answer

  14. SolomonZelman
    • one year ago
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    I am doing a suicide mission

  15. xapproachesinfinity
    • one year ago
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    2nd diverge actually! just limit

  16. SolomonZelman
    • one year ago
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    i am getting stuck on the first one

  17. SolomonZelman
    • one year ago
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    \(\Large\color{black}{ \displaystyle \lim_{ n\rightarrow \infty }\frac{(n+1)!~^{n+1}n^{(n!)}}{(n+1)^{(n+1)!}(n!)^n}}\) this is the first outcome of the ratio test a(n+1)/a(n)

  18. xapproachesinfinity
    • one year ago
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    you can kill (n!)^n leaving (n+1)^(n+1)

  19. SolomonZelman
    • one year ago
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    oh...

  20. SolomonZelman
    • one year ago
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    indeed... i didn't notice that... to complicated it got

  21. xapproachesinfinity
    • one year ago
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    (n+1)^(n+1)!=((n+1)^(n+1))^n!

  22. SolomonZelman
    • one year ago
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    wow.... \(\bf\color{red}{\href{http:///www.wolframalpha.com/input/?i=limit+n%E2%86%92%E2%88%9E+%28%28%28n%2B1%29%21%29%5E%28n%2B1%29n%5E%28n%21%29%29%2F%28%28n%2B1%29%5E%28%28n%2B1%29%21%29%28n%21%29%5En%29}{RESULT ~is~0}}\)

  23. xapproachesinfinity
    • one year ago
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    eh really?

  24. SolomonZelman
    • one year ago
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    yes converges....

  25. SolomonZelman
    • one year ago
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    i entered the limit, r=0

  26. SolomonZelman
    • one year ago
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    as n-> inf

  27. xapproachesinfinity
    • one year ago
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    in computer program? or you did the limit

  28. SolomonZelman
    • one year ago
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    i entered in wolfram and did a hiperlink, because the link is kinda too long

  29. xapproachesinfinity
    • one year ago
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    oh! i think we can do it by hand lolz needs some good cleaning

  30. anonymous
    • one year ago
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    hello solomon you have the coolest profile pic

  31. anonymous
    • one year ago
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    can i use it please?

  32. SolomonZelman
    • one year ago
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    plagiarism.... i would mind that someone looks same as me. do some basic effects to it and change it so that it differs

  33. SolomonZelman
    • one year ago
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    anyway....

  34. xapproachesinfinity
    • one year ago
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    we can use some taylor to find out but that's too long of work the limit (n+1)^n+1=1

  35. SolomonZelman
    • one year ago
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    i got disconnected.

  36. SolomonZelman
    • one year ago
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    yes, that is not something i would right now, i am a bit tired after finals

  37. SolomonZelman
    • one year ago
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    I will do some more very simple ones I guess. tnx:)

  38. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}}\)

  39. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle e^x=\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\) \(\large\color{black}{ \displaystyle ee^x=e\sum_{ n=1 }^{ \infty } \frac{x^n}{n!}}\) \(\large\color{black}{ \displaystyle ee^1=e\sum_{ n=1 }^{ \infty } \frac{1^n}{n!}}\) \(\large\color{black}{ \displaystyle e^2=\sum_{ n=1 }^{ \infty } \frac{e}{n!}}\)

  40. SolomonZelman
    • one year ago
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    i have an error

  41. SolomonZelman
    • one year ago
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    this x^n/n! starts from 0

  42. xapproachesinfinity
    • one year ago
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    that's pretty good trick :) you can make that start at 1 instead 3

  43. SolomonZelman
    • one year ago
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    and then \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-\frac{e}{0!}-\frac{e}{1!}-\frac{e}{2!} }\) \(\large\color{black}{ \displaystyle \sum_{ n=3 }^{ \infty } \frac{e}{n!}=e^2-2.5e }\)

  44. SolomonZelman
    • one year ago
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    yes, as long as I change my both sides the same way it can all work out...

  45. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } \frac{n}{n!} }\) for convr/diver

  46. SolomonZelman
    • one year ago
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    i mean that is obviously convergent

  47. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle (n+1)/(n+1)!~~\times ~~~n!/n}\) \(\large\color{black}{ \displaystyle1/n=0}\)

  48. SolomonZelman
    • one year ago
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    :)

  49. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{ n=k }^{ \infty } \frac{n!}{n^n}}\)

  50. xapproachesinfinity
    • one year ago
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    i have seen that one before i guess

  51. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle n!=1\times 2\times 3 \times ... \times n}\) \(\large\color{black}{ \displaystyle n^n=n\times n\times n \times ... \times n}\)

  52. SolomonZelman
    • one year ago
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    oh,comp test to 2/n^2

  53. xapproachesinfinity
    • one year ago
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    ratio works good with that too

  54. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \frac{n!}{n^n}=\frac{1}{n}\times \frac{2}{n} \times \frac{3}{n} \times \frac{4}{n} \times ~~...~~\times \frac{n}{n}}\) (n times) each of these terms fraction on the right is smaller than 1 besides from 1/n so n!/n^n is in fact smaller than the product of 1st 2 terms

  55. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{n=k}^{\infty}\frac{n!}{n^n}=\sum_{n=k}^{\infty}\frac{2}{n^2}}\)

  56. SolomonZelman
    • one year ago
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    by the way what is the p series 1/n^2 ? 6/pi^2 ?

  57. xapproachesinfinity
    • one year ago
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    1/n^p

  58. xapproachesinfinity
    • one year ago
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    constant on top doesnot really matter

  59. xapproachesinfinity
    • one year ago
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    oh what is it yes pi^2/6

  60. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2/6}\)

  61. SolomonZelman
    • one year ago
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    and with p=4, pi^4/90

  62. SolomonZelman
    • one year ago
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    starting from n=1, and w/o coefficients in front

  63. SolomonZelman
    • one year ago
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    k, lets see what math math got there:)

  64. xapproachesinfinity
    • one year ago
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    well you are just comparing there are not equal

  65. xapproachesinfinity
    • one year ago
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    you want to find the limit?

  66. SolomonZelman
    • one year ago
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    they are not equal, but 2/n^2 is larger than n!/n^n

  67. SolomonZelman
    • one year ago
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    |dw:1435010612413:dw|

  68. SolomonZelman
    • one year ago
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    |dw:1435010912898:dw|

  69. SolomonZelman
    • one year ago
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    when you multiply something by a number smaller than 1 you make it smaller

  70. SolomonZelman
    • one year ago
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    this is why 2/n^2 is greater than n!/n^n

  71. SolomonZelman
    • one year ago
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    so call n!/n^n A(n) and 2/n^2 B(n)

  72. SolomonZelman
    • one year ago
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    by definition of comparison test, if B(n) is larger and converges, then A(n) will also converges

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