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Prove that $forall m in mathbb{N}^{*},exists n in mathbb{N},forall k geq n, p_{k+1}^m

1

$begingroup$

I need to prove :

$$forall m in mathbb{N}^{*},exists n in mathbb{N},forall k geq n, p_{k+1}^m<prod_{i=1}^{k}p_i$$

I can prove this assertion using Prime number theorem :

For fixed $m$ we have : $logleft(displaystyle prod_{i=1}^{k}p_i right) sim p_k$ and $log(p_{k+1}) sim log(k+1)$ give the result.

But I need another proof not asymptotic.

number-theory elementary-number-theory prime-numbers arithmetic

share|cite|improve this question

edited Mar 22 at 18:35

LAGRIDA

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bertrand postulate?
  $endgroup$
  – Collag3n
  Mar 22 at 10:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Betrand postulate gives $p_i < p_{i+1} < 2p_i$. But the question is about $p_{k+1}$ and the primorial of $p_k$
  $endgroup$
  – LAGRIDA
  Mar 22 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bertrand postulate gives $p_{i} < 2^{i} implies p_{k+1}^m < 2^{m(k+1)}$ and if we can proove $2^{m(k+1)} < displaystyle prod_{i=1}^{k}p_i$ for some $k geq n$
  $endgroup$
  – LAGRIDA
  Mar 22 at 11:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have $2^{leftlfloor log(p_i)/log(2) rightrfloor} leq p_i$
  $endgroup$
  – LAGRIDA
  Mar 22 at 11:15
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can someone explain why the downvote
  $endgroup$
  – HAMIDINE SOUMARE
  Mar 22 at 14:18
  

  
  
  
  

 | 
show 5 more comments

1

$begingroup$

I need to prove :

$$forall m in mathbb{N}^{*},exists n in mathbb{N},forall k geq n, p_{k+1}^m<prod_{i=1}^{k}p_i$$

I can prove this assertion using Prime number theorem :

For fixed $m$ we have : $logleft(displaystyle prod_{i=1}^{k}p_i right) sim p_k$ and $log(p_{k+1}) sim log(k+1)$ give the result.

But I need another proof not asymptotic.

number-theory elementary-number-theory prime-numbers arithmetic

share|cite|improve this question

edited Mar 22 at 18:35

LAGRIDA

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bertrand postulate?
  $endgroup$
  – Collag3n
  Mar 22 at 10:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Betrand postulate gives $p_i < p_{i+1} < 2p_i$. But the question is about $p_{k+1}$ and the primorial of $p_k$
  $endgroup$
  – LAGRIDA
  Mar 22 at 10:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Bertrand postulate gives $p_{i} < 2^{i} implies p_{k+1}^m < 2^{m(k+1)}$ and if we can proove $2^{m(k+1)} < displaystyle prod_{i=1}^{k}p_i$ for some $k geq n$
  $endgroup$
  – LAGRIDA
  Mar 22 at 11:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We have $2^{leftlfloor log(p_i)/log(2) rightrfloor} leq p_i$
  $endgroup$
  – LAGRIDA
  Mar 22 at 11:15
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Can someone explain why the downvote
  $endgroup$
  – HAMIDINE SOUMARE
  Mar 22 at 14:18
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

I need to prove :

$$forall m in mathbb{N}^{*},exists n in mathbb{N},forall k geq n, p_{k+1}^m<prod_{i=1}^{k}p_i$$

I can prove this assertion using Prime number theorem :

For fixed $m$ we have : $logleft(displaystyle prod_{i=1}^{k}p_i right) sim p_k$ and $log(p_{k+1}) sim log(k+1)$ give the result.

But I need another proof not asymptotic.

number-theory elementary-number-theory prime-numbers arithmetic

share|cite|improve this question

edited Mar 22 at 18:35

LAGRIDA

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113

$endgroup$

share|cite|improve this question

edited Mar 22 at 18:35

LAGRIDA

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113

share|cite|improve this question

edited Mar 22 at 18:35

LAGRIDA

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113

asked Mar 22 at 8:23

LAGRIDALAGRIDA

280113