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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
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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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3this is an induction proof, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I am not sure about that, but if it helps, why not, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3really 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:)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hey, 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?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1And the book said Bertrand could not produce a proof \(\implies\) I cannot either. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@IrishBoy123 , please, tell me dw:1443050900999:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3i would say does not converge

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"really sorry loser i am crap at this pure math stuff" < sameeeeeeeeeeeeee

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0can we take that as an infinite GP with common ratio=1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Since I have to prove \(sin (z) = \sum_{k=0}^\infty \dfrac{(1)^k z^{2k+1}}{(2k+1)!}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I end up with that i but cannot take it down to (1)^k

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Let me post my proof. \(sin z = \dfrac{1}{2}(e^{iz}e^{iz} =\dfrac{1}{2}\sum_{n=0}^\infty (\dfrac{(iz)^n}{n!}\dfrac{(iz)^n}{n!})\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1factor i^n z^n out, we get \(sin z = (1/2) \sum_{n=0}^\infty (1(1)^n) (\dfrac{i^nz^n}{n!})\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\[i + i^3 + i^5 + i^7 + ...\] \[= i( 1+ i^2 + i^4 + i^6 + ...)\] \[= i( 1+(1) + (1) + (1) + ...)\] does not converge

Loser66
 one year ago
Best ResponseYou've already chosen the best response.11(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 \(sin z = \sum_{k=0}^\infty \dfrac{i^{2k+1}z^{2k+1}}{(2k+1)!}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Now, 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. :(

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3\( i^{(2k+1)} = i^{(2k)}*i = (1)^k *i\) yes

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, I got my mistake!! \(sin z = \dfrac{e^{iz}e^{iz}}{2i}\) that cancel out with my i.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thanks a lot, friends. Thanks for being here so that I am not alone with my problem.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3you mean \[sin z = \dfrac{1}{2\color{red}i}(e^{iz}e^{iz} =\dfrac{1}{2\color{red}i}\sum_{n=0}^\infty (\dfrac{(iz)^n}{n!}\dfrac{(iz)^n}{n!})\]
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