Proving $left|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq 2q |s| (m/x)^{sigma}$
$text{Show}:displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq 2q |s| left(frac mxright)^{sigma}$ where $s=sigma +it$.
Here is what I tried:
$displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq left(frac mxright)^{sigma} sum_{n>x/m} left| chi(n) right|$.
I dont know how to get the $2qs$ part. Here, we assume $chi$ is a character $mod q$.
analytic-number-theory
edited Nov 23 at 21:14
Yadati Kiran
1,702519
asked Nov 23 at 20:17
usere5225321
617412
add a comment |
$text{Show}:displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq 2q |s| left(frac mxright)^{sigma}$ where $s=sigma +it$.
Here is what I tried:
$displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq left(frac mxright)^{sigma} sum_{n>x/m} left| chi(n) right|$.
I dont know how to get the $2qs$ part. Here, we assume $chi$ is a character $mod q$.
analytic-number-theory
edited Nov 23 at 21:14
Yadati Kiran
1,702519
asked Nov 23 at 20:17
usere5225321
617412
1
$sum_{n=N}^infty chi(n) n^{-s} = sum_{n=N}^infty A_chi(n) (n^{-s}-(n+1)^{-s})$ where $A_chi(n) = sum_{m=N}^n chi(m)$
– reuns
Nov 23 at 21:19
add a comment |
$text{Show}:displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq 2q |s| left(frac mxright)^{sigma}$ where $s=sigma +it$.
Here is what I tried:
$displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq left(frac mxright)^{sigma} sum_{n>x/m} left| chi(n) right|$.
I dont know how to get the $2qs$ part. Here, we assume $chi$ is a character $mod q$.
analytic-number-theory
edited Nov 23 at 21:14
Yadati Kiran
1,702519
asked Nov 23 at 20:17
usere5225321
617412
$text{Show}:displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq 2q |s| left(frac mxright)^{sigma}$ where $s=sigma +it$.
Here is what I tried:
$displaystyleleft|sum_{n>x/m} frac{ chi(n)}{ n^{s}} right| leq left(frac mxright)^{sigma} sum_{n>x/m} left| chi(n) right|$.
I dont know how to get the $2qs$ part. Here, we assume $chi$ is a character $mod q$.
analytic-number-theory
analytic-number-theory
edited Nov 23 at 21:14
Yadati Kiran
1,702519
asked Nov 23 at 20:17
usere5225321
617412
edited Nov 23 at 21:14
Yadati Kiran
1,702519
asked Nov 23 at 20:17
usere5225321
617412
edited Nov 23 at 21:14
Yadati Kiran
1,702519
edited Nov 23 at 21:14
Yadati Kiran
1,702519
edited Nov 23 at 21:14
Yadati Kiran
1,702519
1,702519
asked Nov 23 at 20:17
usere5225321
617412
asked Nov 23 at 20:17
usere5225321
617412
asked Nov 23 at 20:17
usere5225321
617412
617412
1
$sum_{n=N}^infty chi(n) n^{-s} = sum_{n=N}^infty A_chi(n) (n^{-s}-(n+1)^{-s})$ where $A_chi(n) = sum_{m=N}^n chi(m)$
– reuns
Nov 23 at 21:19
add a comment |
1
$sum_{n=N}^infty chi(n) n^{-s} = sum_{n=N}^infty A_chi(n) (n^{-s}-(n+1)^{-s})$ where $A_chi(n) = sum_{m=N}^n chi(m)$
– reuns
Nov 23 at 21:19
1
1
$sum_{n=N}^infty chi(n) n^{-s} = sum_{n=N}^infty A_chi(n) (n^{-s}-(n+1)^{-s})$ where $A_chi(n) = sum_{m=N}^n chi(m)$
– reuns
Nov 23 at 21:19
$sum_{n=N}^infty chi(n) n^{-s} = sum_{n=N}^infty A_chi(n) (n^{-s}-(n+1)^{-s})$ where $A_chi(n) = sum_{m=N}^n chi(m)$
– reuns
Nov 23 at 21:19
add a comment |
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1
$sum_{n=N}^infty chi(n) n^{-s} = sum_{n=N}^infty A_chi(n) (n^{-s}-(n+1)^{-s})$ where $A_chi(n) = sum_{m=N}^n chi(m)$
– reuns
Nov 23 at 21:19