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Prove that $forall m in mathbbN^*,exists n in mathbbN,forall k geq n, p_k+1^m


Is it true that $sum_k=1^n(p_kprod_i=1^k(1-p_i)) stackrelmbox?= 1 - prod_i=1^n(1-p_i)$Estimate for the product of primes less than nProve or disprove that $forall kinmathbb N$ there exist tree consecutive primes such that $p_i-p_i-1gt k$ and $p_i+1-p_igt k$Inequality with Euler's totient functionProve if $n=p_1p_2cdots p_k +1$, then for every $i$, $i=1,2,cdots k, p_i$ does not divide n.Prove that $prod_i<j (p_i^p_j-p_j^p_i)$ is divisible by $5777$Prove $prod_c < p leq x left(1 - fraccp right) ll log ^-c x .$Prove that $forall xinmathbbN text there always exists a prime pequiv1 pmod 6 text s.t. p|(2x)^2+3;$Interesting missing proof (Prime Number Theorem)Weak k-Tuple conjecture form and what we should proof













1












$begingroup$


I need to prove :



$$forall m in mathbbN^*,exists n in mathbbN,forall k geq n, p_k+1^m<prod_i=1^kp_i$$



I can prove this assertion using Prime number theorem :



For fixed $m$ we have : $logleft(displaystyle prod_i=1^kp_i right) sim p_k$ and $log(p_k+1) sim log(k+1)$ give the result.



But I need another proof not asymptotic.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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 $2^m(k+1) < displaystyle prod_i=1^kp_i$ 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















1












$begingroup$


I need to prove :



$$forall m in mathbbN^*,exists n in mathbbN,forall k geq n, p_k+1^m<prod_i=1^kp_i$$



I can prove this assertion using Prime number theorem :



For fixed $m$ we have : $logleft(displaystyle prod_i=1^kp_i right) sim p_k$ and $log(p_k+1) sim log(k+1)$ give the result.



But I need another proof not asymptotic.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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 $2^m(k+1) < displaystyle prod_i=1^kp_i$ 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













1












1








1





$begingroup$


I need to prove :



$$forall m in mathbbN^*,exists n in mathbbN,forall k geq n, p_k+1^m<prod_i=1^kp_i$$



I can prove this assertion using Prime number theorem :



For fixed $m$ we have : $logleft(displaystyle prod_i=1^kp_i right) sim p_k$ and $log(p_k+1) sim log(k+1)$ give the result.



But I need another proof not asymptotic.










share|cite|improve this question











$endgroup$




I need to prove :



$$forall m in mathbbN^*,exists n in mathbbN,forall k geq n, p_k+1^m<prod_i=1^kp_i$$



I can prove this assertion using Prime number theorem :



For fixed $m$ we have : $logleft(displaystyle prod_i=1^kp_i right) sim p_k$ 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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 22 at 18:35







LAGRIDA

















asked Mar 22 at 8:23









LAGRIDALAGRIDA

270112




270112











  • $begingroup$
    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 $2^m(k+1) < displaystyle prod_i=1^kp_i$ 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
















  • $begingroup$
    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 $2^m(k+1) < displaystyle prod_i=1^kp_i$ 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















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




$begingroup$
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




$begingroup$
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 $2^m(k+1) < displaystyle prod_i=1^kp_i$ for some $k geq n$
$endgroup$
– LAGRIDA
Mar 22 at 11:14




$begingroup$
Bertrand postulate gives $p_i < 2^i implies p_k+1^m < 2^m(k+1)$ and if we can proove $2^m(k+1) < displaystyle prod_i=1^kp_i$ 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





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





1




1




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




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










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