## Loser66 one year ago Show that for all integer n >0, there exists a prime p such that n < p =< n!+1 Please, help

1. IrishBoy123

this is an induction proof, right?

2. Loser66

I am not sure about that, but if it helps, why not, right?

3. Loser66

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

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

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

And the book said Bertrand could not produce a proof $$\implies$$ I cannot either. :)

7. Loser66

@dan815

8. Loser66

@amistre64

9. Loser66

@IrishBoy123 , please, tell me |dw:1443050900999:dw|

10. Loser66

|dw:1443050992998:dw|

11. imqwerty

i/2

12. IrishBoy123

i would say does not converge

13. anonymous

"really sorry loser i am crap at this pure math stuff" <-- sameeeeeeeeeeeeee

14. imqwerty

can we take that as an infinite GP with common ratio=-1

15. Loser66

|dw:1443051096482:dw|

16. Loser66

Since I have to prove $$sin (z) = \sum_{k=0}^\infty \dfrac{(-1)^k z^{2k+1}}{(2k+1)!}$$

17. Loser66

I end up with that i but cannot take it down to (-1)^k

18. Loser66

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

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

$i + i^3 + i^5 + i^7 + ...$ $= i( 1+ i^2 + i^4 + i^6 + ...)$ $= i( 1+(-1) + (1) + (-1) + ...)$ does not converge

21. Loser66

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

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

$$i^{(2k+1)} = i^{(2k)}*i = (-1)^k *i$$ yes

24. Loser66

oh, I got my mistake!! $$sin z = \dfrac{e^{iz}-e^{-iz}}{2i}$$ that cancel out with my i.

25. Loser66

Thanks a lot, friends. Thanks for being here so that I am not alone with my problem.

26. imqwerty

hehe :D

27. IrishBoy123

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

yes