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Loser66

  • one year ago

Show that for all integer n >0, there exists a prime p such that n < p =< n!+1 Please, help

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  1. IrishBoy123
    • one year ago
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    this is an induction proof, right?

  2. Loser66
    • one year ago
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    I am not sure about that, but if it helps, why not, right?

  3. Loser66
    • one year ago
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    I can prove the LHS and the middle like Let p is an arbitrary prime, then (p, 1) =1. By Bezouth theorem, in the sequence \(\{pn+1\}_{n=1}^\infty \) there are infinite many prime, hence we always have a prime under that form pn +1 Surely, with integer n>0, n < pn +1. But not sure about the far right.

  4. IrishBoy123
    • one year ago
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    really sorry loser, i am crap at this pure math stuff; but i did find Bertrand's Postulate: for every \(n > 1\), here is always at least one prime p such that: \[n < p < 2n\] interesting stuff:-)

  5. Loser66
    • one year ago
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    Hey, is it not that it is done? since 2n < n! = 1*2*3*....*n <n! +1 Wow!! unfortunately, if I use it, I have to prove Bertrand's postulate, right?

  6. Loser66
    • one year ago
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    And the book said Bertrand could not produce a proof \(\implies\) I cannot either. :)

  7. Loser66
    • one year ago
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    @dan815

  8. Loser66
    • one year ago
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    @amistre64

  9. Loser66
    • one year ago
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    @IrishBoy123 , please, tell me |dw:1443050900999:dw|

  10. Loser66
    • one year ago
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    |dw:1443050992998:dw|

  11. imqwerty
    • one year ago
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    i/2

  12. IrishBoy123
    • one year ago
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    i would say does not converge

  13. anonymous
    • one year ago
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    "really sorry loser i am crap at this pure math stuff" <-- sameeeeeeeeeeeeee

  14. imqwerty
    • one year ago
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    can we take that as an infinite GP with common ratio=-1

  15. Loser66
    • one year ago
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    |dw:1443051096482:dw|

  16. Loser66
    • one year ago
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    Since I have to prove \(sin (z) = \sum_{k=0}^\infty \dfrac{(-1)^k z^{2k+1}}{(2k+1)!}\)

  17. Loser66
    • one year ago
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    I end up with that i but cannot take it down to (-1)^k

  18. Loser66
    • one year ago
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    Let me post my proof. \(sin z = \dfrac{1}{2}(e^{iz}-e^{-iz} =\dfrac{1}{2}\sum_{n=0}^\infty (\dfrac{(iz)^n}{n!}-\dfrac{(-iz)^n}{n!})\)

  19. Loser66
    • one year ago
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    factor i^n z^n out, we get \(sin z = (1/2) \sum_{n=0}^\infty (1-(-1)^n) (\dfrac{i^nz^n}{n!})\)

  20. IrishBoy123
    • one year ago
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    \[i + i^3 + i^5 + i^7 + ...\] \[= i( 1+ i^2 + i^4 + i^6 + ...)\] \[= i( 1+(-1) + (1) + (-1) + ...)\] does not converge

  21. Loser66
    • one year ago
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    1-(-1)^n = o if n even 1-(-1) ^n =2 if n odd hence, we take if n odd only, so that n = 2k +1 replace all \(sin z = \sum_{k=0}^\infty \dfrac{i^{2k+1}z^{2k+1}}{(2k+1)!}\)

  22. Loser66
    • one year ago
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    Now, my question on i^(2k+1) = i^(2k)*i = (-1)^k *i I can't get rid of that i. so that the proof is not done. :(

  23. IrishBoy123
    • one year ago
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    \( i^{(2k+1)} = i^{(2k)}*i = (-1)^k *i\) yes

  24. Loser66
    • one year ago
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    oh, I got my mistake!! \(sin z = \dfrac{e^{iz}-e^{-iz}}{2i}\) that cancel out with my i.

  25. Loser66
    • one year ago
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    Thanks a lot, friends. Thanks for being here so that I am not alone with my problem.

  26. imqwerty
    • one year ago
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    hehe :D

  27. IrishBoy123
    • one year ago
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    you mean \[sin z = \dfrac{1}{2\color{red}i}(e^{iz}-e^{-iz} =\dfrac{1}{2\color{red}i}\sum_{n=0}^\infty (\dfrac{(iz)^n}{n!}-\dfrac{(-iz)^n}{n!})\]

  28. Loser66
    • one year ago
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    yes

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