$zeta(1+it)=0$ 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.
analytic-number-theory
asked Nov 20 at 19:47
usere5225321
592311
add a comment |
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.
analytic-number-theory
asked Nov 20 at 19:47
usere5225321
592311
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
add a comment |
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.
analytic-number-theory
asked Nov 20 at 19:47
usere5225321
592311
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
analytic-number-theory
asked Nov 20 at 19:47
usere5225321
592311
asked Nov 20 at 19:47
usere5225321
592311
edited Nov 20 at 20:09
asked Nov 20 at 19:47
usere5225321
592311
asked Nov 20 at 19:47
usere5225321
592311
asked Nov 20 at 19:47
usere5225321
592311
592311
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
add a comment |
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
add a comment |
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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