$zeta(1+it)=0$ implies sum of prime if finite?











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Can anyone explain why $zeta(1+it)= sum_{n=1}^{infty} frac{1}{n^{1+it}}=0$ will imply $sum_{text{p prime}} frac{1-Re(-p^{-it})}{p}< infty?$



Here is what I have so far:



We know $zeta(s)$ does not have any zeros for $Re(s) geq 1.$ So $zeta(1+it) not=0.$



Thus, we have,



$sum_{text{p prime}}frac{1-Re(-p^{-it})}{p}= sum_{text{p prime}} frac{1}{p}- sum_{text{p prime}}frac{Re(-p^{-it})}{p}$.



The first sum is known to be infinite and the second one ressembles $zeta(1+it).$ However, I dont know how to conclude the whole sum is infinite.










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  • It's surely because $zeta(1+it)$ is never zero.
    – Lord Shark the Unknown
    Nov 20 at 19:58










  • @LordSharktheUnknown question edited !
    – usere5225321
    Nov 20 at 20:09















up vote
0
down vote

favorite












Can anyone explain why $zeta(1+it)= sum_{n=1}^{infty} frac{1}{n^{1+it}}=0$ will imply $sum_{text{p prime}} frac{1-Re(-p^{-it})}{p}< infty?$



Here is what I have so far:



We know $zeta(s)$ does not have any zeros for $Re(s) geq 1.$ So $zeta(1+it) not=0.$



Thus, we have,



$sum_{text{p prime}}frac{1-Re(-p^{-it})}{p}= sum_{text{p prime}} frac{1}{p}- sum_{text{p prime}}frac{Re(-p^{-it})}{p}$.



The first sum is known to be infinite and the second one ressembles $zeta(1+it).$ However, I dont know how to conclude the whole sum is infinite.










share|cite|improve this question
























  • It's surely because $zeta(1+it)$ is never zero.
    – Lord Shark the Unknown
    Nov 20 at 19:58










  • @LordSharktheUnknown question edited !
    – usere5225321
    Nov 20 at 20:09













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Can anyone explain why $zeta(1+it)= sum_{n=1}^{infty} frac{1}{n^{1+it}}=0$ will imply $sum_{text{p prime}} frac{1-Re(-p^{-it})}{p}< infty?$



Here is what I have so far:



We know $zeta(s)$ does not have any zeros for $Re(s) geq 1.$ So $zeta(1+it) not=0.$



Thus, we have,



$sum_{text{p prime}}frac{1-Re(-p^{-it})}{p}= sum_{text{p prime}} frac{1}{p}- sum_{text{p prime}}frac{Re(-p^{-it})}{p}$.



The first sum is known to be infinite and the second one ressembles $zeta(1+it).$ However, I dont know how to conclude the whole sum is infinite.










share|cite|improve this question















Can anyone explain why $zeta(1+it)= sum_{n=1}^{infty} frac{1}{n^{1+it}}=0$ will imply $sum_{text{p prime}} frac{1-Re(-p^{-it})}{p}< infty?$



Here is what I have so far:



We know $zeta(s)$ does not have any zeros for $Re(s) geq 1.$ So $zeta(1+it) not=0.$



Thus, we have,



$sum_{text{p prime}}frac{1-Re(-p^{-it})}{p}= sum_{text{p prime}} frac{1}{p}- sum_{text{p prime}}frac{Re(-p^{-it})}{p}$.



The first sum is known to be infinite and the second one ressembles $zeta(1+it).$ However, I dont know how to conclude the whole sum is infinite.







analytic-number-theory






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edited Nov 20 at 20:09

























asked Nov 20 at 19:47









usere5225321

612412




612412












  • It's surely because $zeta(1+it)$ is never zero.
    – Lord Shark the Unknown
    Nov 20 at 19:58










  • @LordSharktheUnknown question edited !
    – usere5225321
    Nov 20 at 20:09


















  • It's surely because $zeta(1+it)$ is never zero.
    – Lord Shark the Unknown
    Nov 20 at 19:58










  • @LordSharktheUnknown question edited !
    – usere5225321
    Nov 20 at 20:09
















It's surely because $zeta(1+it)$ is never zero.
– Lord Shark the Unknown
Nov 20 at 19:58




It's surely because $zeta(1+it)$ is never zero.
– Lord Shark the Unknown
Nov 20 at 19:58












@LordSharktheUnknown question edited !
– usere5225321
Nov 20 at 20:09




@LordSharktheUnknown question edited !
– usere5225321
Nov 20 at 20:09















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