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Is this $zeta(p)=sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1$ true with sigma is power of sum divisor function?

Sum of the unitary divisor function (mod p)Looking for series representations of Riemann's zeta function valid for $sigma<1$.Please help me understand Analytic Density $lim_sigma to 1^+frac1zeta(sigma)sum_n in A frac1n^sigma$Is this true :$zeta(s-1)=sum_n=1^inftyfracgcd(n,n)operatornamelcm(n,n)^s$?Is right this application of Hadamard three-lines theorem for $ fraczeta(s)s- fracdzeta(s)dsigma$?order of $sum_k geq 2 fracLambda(k - 1)k^sigma + it$ as $t$ goes to infinityPower Summation Over the Divisor FunctionOn $fracsigma(n+1)p_n+1-fracsigma(n)p_n$, where $p_k$ is the $kth$ prime number and $sigma(k)$ the sum of divisors functionCombinatorial Interpretation of the Sum-of-Divisor Function?What is this sum equal to? $sigma(n)=sum_ineq j frac1i^n j^n$

1

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let $sigma_k(n) =sum_n d^k$ is a sum of divisor function , And let $sigma_k(n)$ be the iterating divisor function, We have for every prime $p$ and for every integer $kgeq 1$ : $$sigma_k(p)=p^k+1 tag1$$ , Now form $(1)$ we have :$$sum_kgeq1 ,pin mathbbPp^-k= sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1tag2$$, then :
$$zeta(p)=sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1tag3$$ , Now my problem if i want to calculate for example $zeta(2)$ I will get $1$ by $(3)$ ? then why that's not true however i have used correct formula ?

prime-numbers riemann-zeta divisor-sum

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asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133

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  1
  

  
  
  
  
  
  

  
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  Your (3) is a nonsense. The Euler product is $log zeta(s)=-sum_p log(1-p^-s)= sum_p^k fracp^-skk$ where in number theory $sum_p, sum_p^k$ means sum over primes and prime powers
  $endgroup$
  – reuns
  Mar 20 at 18:06
  
  

  
  
  
  


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  So what is the $p$ in the LHS of (3) ? (The title contains the same problem.)
  $endgroup$
  – dan_fulea
  Mar 20 at 18:27
  

  
  
  
  


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  $begingroup$
  is a prime number
  $endgroup$
  – zeraoulia rafik
  Mar 20 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, does the RHS depend on this one fixed prime number $p$? Could we use some other letter for the running letter in the undefined set $Bbb P$?
  $endgroup$
  – dan_fulea
  Mar 20 at 18:31
  

  
  
  
  

add a comment | 

1

$begingroup$

let $sigma_k(n) =sum_n d^k$ is a sum of divisor function , And let $sigma_k(n)$ be the iterating divisor function, We have for every prime $p$ and for every integer $kgeq 1$ : $$sigma_k(p)=p^k+1 tag1$$ , Now form $(1)$ we have :$$sum_kgeq1 ,pin mathbbPp^-k= sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1tag2$$, then :
$$zeta(p)=sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1tag3$$ , Now my problem if i want to calculate for example $zeta(2)$ I will get $1$ by $(3)$ ? then why that's not true however i have used correct formula ?

prime-numbers riemann-zeta divisor-sum

share|cite|improve this question

asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your (3) is a nonsense. The Euler product is $log zeta(s)=-sum_p log(1-p^-s)= sum_p^k fracp^-skk$ where in number theory $sum_p, sum_p^k$ means sum over primes and prime powers
  $endgroup$
  – reuns
  Mar 20 at 18:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So what is the $p$ in the LHS of (3) ? (The title contains the same problem.)
  $endgroup$
  – dan_fulea
  Mar 20 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  is a prime number
  $endgroup$
  – zeraoulia rafik
  Mar 20 at 18:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK, does the RHS depend on this one fixed prime number $p$? Could we use some other letter for the running letter in the undefined set $Bbb P$?
  $endgroup$
  – dan_fulea
  Mar 20 at 18:31
  

  
  
  
  

add a comment | 

$begingroup$

let $sigma_k(n) =sum_n d^k$ is a sum of divisor function , And let $sigma_k(n)$ be the iterating divisor function, We have for every prime $p$ and for every integer $kgeq 1$ : $$sigma_k(p)=p^k+1 tag1$$ , Now form $(1)$ we have :$$sum_kgeq1 ,pin mathbbPp^-k= sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1tag2$$, then :
$$zeta(p)=sum_kgeq1 ,pin mathbbPfrac1sigma_k(p)-1tag3$$ , Now my problem if i want to calculate for example $zeta(2)$ I will get $1$ by $(3)$ ? then why that's not true however i have used correct formula ?

prime-numbers riemann-zeta divisor-sum

share|cite|improve this question

asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133

$endgroup$

share|cite|improve this question

asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133

share|cite|improve this question

asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133

asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133

asked Mar 20 at 18:00

zeraoulia rafikzeraoulia rafik

2,38111133