Mixing
Error in expression of incomplete zeta function?
Background & Question
I realised I could do a different manipulation from the one I did over here Strange method to obtain strange number theoretic identities? However after making some calculations I seems to be getting the wrong answer by perhaps an extra term.
I got the expression:
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
But as $n to infty$ we know the $(frac{1}{zeta(k)})' to 0 $ when $k$ to $1$.
Is the correct answer below?
$$sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} $$
If so where did I go wrong in my proof? And what is the "normal" proof?
Proof
Let us write a relation of the Euler–Mascheroni constant for large $n$.
$$ 1+ frac{1}{2} + frac{1}{3} + frac{1}{4} + dots +frac{1}{n!} sim gamma + ln(n!)$$
Or, multiplying $1/2$ boths sides and let $n! to n!/2$
$$ 0 + frac{1}{2} + 0 + frac{1}{4} + dots +frac{1}{n!} sim frac{gamma}{2} + frac{1}{2}ln(frac{n!}{2})$$
Or, multiplying $1/3$ boths sides and let $n! to n!/3$
$$ 0 + 0 + frac{1}{3} + 0 + dots +frac{1}{n!} sim frac{gamma}{3} + frac{1}{3}ln(frac{n!}{3})$$
And so on $n$ times ... Now multiplying the $r$'th row with $a_r$ and adding vertically (whilst defining $b_r$):
$$ a_1+ frac{a_1}{2} + frac{a_1}{3} + frac{a_1}{4} + dots +frac{a_1}{n!} sim a_1 gamma + a_1 ln(n!)$$
$$ 0 + frac{a_2}{2} + 0 + frac{a_2}{4} + dots +frac{a_2}{n!} sim a_2 frac{gamma}{2} + frac{a_2}{2}ln(frac{n!}{2})$$
$$ 0 + 0 + frac{a_3}{3} + 0 + dots +frac{a_3}{n!} sim a_3 frac{gamma}{3} + frac{a_3}{3}ln(frac{n!}{3})$$
$$vdots $$
$+$
$-----------------------------------$
$$ underbrace{frac{b_1}{1}}_{a_1/1} + underbrace{frac{b_2}{2}}_{(a_1+ a_2)/2} + underbrace{frac{b_3}{3}}_{(a_1+ a_3)/3} + dots sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
In the above we define:
$$b_r = sum_{r|l} a_l text{ }forall text{ } 1 leq r leq n$$
$$b_r = sum_{(r-n)|l} a_l text{ }forall text{ } n+1 leq r leq 2n$$
$$b_r = sum_{(r-2n)|l} a_l text{ }forall text{ } 2n+1 leq r leq 3n$$
$$ vdots $$
$$b_r = sum_{(r-(n-1)!)|l} a_l text{ }forall text{ } n(n-1)!- n +1 leq r leq n!$$
Writing the above properly now:
$$ sum_{r=1}^n frac{b_r}{r} + sum_{r=n+1}^{2n} frac{b_{r-n}}{r} + dots+ sum_{r=n!-n+1}^{n!} frac{b_{r-n!+n }}{r} sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
Rearranging the L.H.S and R.H.S:
$$ sum_{r=1}^n sum_{k=0}^{(n-1)! -1 } b_r ( frac{1}{kn+ r}) sim (gamma+ ln n!) sum_{r=1}^n frac{a_r}{r} - sum_{r=1}^n frac{a_r}{r} ln(r) $$
Also note :
$$a_{l}=sum _{dmid l} mu left({frac {l}{d}}right)b_{d} implies frac{partial a_l}{ partial b_d} = mu left({frac {l}{d}}right) $$
where $mu$ is the mobius function. Let us now act $frac{partial }{partial b_d} $ on both sides where $d leq n$:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim (gamma+ ln n!) sum_{r=1}^n frac{mu(frac{r}{d})}{r} - sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
where $mu(lambda)= 0$ where $lambda$ is not an integer. Simplifying some of the terms:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
Let us take $d to 1$
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ 1}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r) $$
Now using some approximations:
$$ 1 + frac{1}{n} sum_{k=1}^{(n-1)! -1 } frac{1}{k} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
Using the first equation with a different $n$:
$$ 1 + frac{gamma + ln(n-1)!}{n} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
combinatorics number-theory proof-verification alternative-proof zeta-functions
asked Nov 20 '18 at 0:42
More Anonymous
34419
add a comment |
Background & Question
I realised I could do a different manipulation from the one I did over here Strange method to obtain strange number theoretic identities? However after making some calculations I seems to be getting the wrong answer by perhaps an extra term.
I got the expression:
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
But as $n to infty$ we know the $(frac{1}{zeta(k)})' to 0 $ when $k$ to $1$.
Is the correct answer below?
$$sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} $$
If so where did I go wrong in my proof? And what is the "normal" proof?
Proof
Let us write a relation of the Euler–Mascheroni constant for large $n$.
$$ 1+ frac{1}{2} + frac{1}{3} + frac{1}{4} + dots +frac{1}{n!} sim gamma + ln(n!)$$
Or, multiplying $1/2$ boths sides and let $n! to n!/2$
$$ 0 + frac{1}{2} + 0 + frac{1}{4} + dots +frac{1}{n!} sim frac{gamma}{2} + frac{1}{2}ln(frac{n!}{2})$$
Or, multiplying $1/3$ boths sides and let $n! to n!/3$
$$ 0 + 0 + frac{1}{3} + 0 + dots +frac{1}{n!} sim frac{gamma}{3} + frac{1}{3}ln(frac{n!}{3})$$
And so on $n$ times ... Now multiplying the $r$'th row with $a_r$ and adding vertically (whilst defining $b_r$):
$$ a_1+ frac{a_1}{2} + frac{a_1}{3} + frac{a_1}{4} + dots +frac{a_1}{n!} sim a_1 gamma + a_1 ln(n!)$$
$$ 0 + frac{a_2}{2} + 0 + frac{a_2}{4} + dots +frac{a_2}{n!} sim a_2 frac{gamma}{2} + frac{a_2}{2}ln(frac{n!}{2})$$
$$ 0 + 0 + frac{a_3}{3} + 0 + dots +frac{a_3}{n!} sim a_3 frac{gamma}{3} + frac{a_3}{3}ln(frac{n!}{3})$$
$$vdots $$
$+$
$-----------------------------------$
$$ underbrace{frac{b_1}{1}}_{a_1/1} + underbrace{frac{b_2}{2}}_{(a_1+ a_2)/2} + underbrace{frac{b_3}{3}}_{(a_1+ a_3)/3} + dots sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
In the above we define:
$$b_r = sum_{r|l} a_l text{ }forall text{ } 1 leq r leq n$$
$$b_r = sum_{(r-n)|l} a_l text{ }forall text{ } n+1 leq r leq 2n$$
$$b_r = sum_{(r-2n)|l} a_l text{ }forall text{ } 2n+1 leq r leq 3n$$
$$ vdots $$
$$b_r = sum_{(r-(n-1)!)|l} a_l text{ }forall text{ } n(n-1)!- n +1 leq r leq n!$$
Writing the above properly now:
$$ sum_{r=1}^n frac{b_r}{r} + sum_{r=n+1}^{2n} frac{b_{r-n}}{r} + dots+ sum_{r=n!-n+1}^{n!} frac{b_{r-n!+n }}{r} sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
Rearranging the L.H.S and R.H.S:
$$ sum_{r=1}^n sum_{k=0}^{(n-1)! -1 } b_r ( frac{1}{kn+ r}) sim (gamma+ ln n!) sum_{r=1}^n frac{a_r}{r} - sum_{r=1}^n frac{a_r}{r} ln(r) $$
Also note :
$$a_{l}=sum _{dmid l} mu left({frac {l}{d}}right)b_{d} implies frac{partial a_l}{ partial b_d} = mu left({frac {l}{d}}right) $$
where $mu$ is the mobius function. Let us now act $frac{partial }{partial b_d} $ on both sides where $d leq n$:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim (gamma+ ln n!) sum_{r=1}^n frac{mu(frac{r}{d})}{r} - sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
where $mu(lambda)= 0$ where $lambda$ is not an integer. Simplifying some of the terms:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
Let us take $d to 1$
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ 1}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r) $$
Now using some approximations:
$$ 1 + frac{1}{n} sum_{k=1}^{(n-1)! -1 } frac{1}{k} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
Using the first equation with a different $n$:
$$ 1 + frac{gamma + ln(n-1)!}{n} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
combinatorics number-theory proof-verification alternative-proof zeta-functions
asked Nov 20 '18 at 0:42
More Anonymous
34419
$sum_{n=1}^infty frac{mu(n)}{n} (-log n)^k$ converges for every $k$ (prime number theorem) and the rate of convergence encodes the zero-free region of $zeta(s)$ (thus the Riemann hypothesis). In particular $sum_{n=1}^N frac{mu(n)}{n} (-log n)^k= frac{d^k}{ds^k} frac{1}{zeta(s)}|_{s=1}+ O(N^{-1+sigma+epsilon})$ iff $zeta(s)$ has no zeros for $Re(s) > sigma$.
– reuns
Nov 20 '18 at 1:04
@reuns So this is a highly non-trivial statement I'm trying to make?
– More Anonymous
Nov 20 '18 at 1:07
I'll be editing this post (I caught a foolish error)
– More Anonymous
Nov 20 '18 at 2:28
add a comment |
Background & Question
I realised I could do a different manipulation from the one I did over here Strange method to obtain strange number theoretic identities? However after making some calculations I seems to be getting the wrong answer by perhaps an extra term.
I got the expression:
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
But as $n to infty$ we know the $(frac{1}{zeta(k)})' to 0 $ when $k$ to $1$.
Is the correct answer below?
$$sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} $$
If so where did I go wrong in my proof? And what is the "normal" proof?
Proof
Let us write a relation of the Euler–Mascheroni constant for large $n$.
$$ 1+ frac{1}{2} + frac{1}{3} + frac{1}{4} + dots +frac{1}{n!} sim gamma + ln(n!)$$
Or, multiplying $1/2$ boths sides and let $n! to n!/2$
$$ 0 + frac{1}{2} + 0 + frac{1}{4} + dots +frac{1}{n!} sim frac{gamma}{2} + frac{1}{2}ln(frac{n!}{2})$$
Or, multiplying $1/3$ boths sides and let $n! to n!/3$
$$ 0 + 0 + frac{1}{3} + 0 + dots +frac{1}{n!} sim frac{gamma}{3} + frac{1}{3}ln(frac{n!}{3})$$
And so on $n$ times ... Now multiplying the $r$'th row with $a_r$ and adding vertically (whilst defining $b_r$):
$$ a_1+ frac{a_1}{2} + frac{a_1}{3} + frac{a_1}{4} + dots +frac{a_1}{n!} sim a_1 gamma + a_1 ln(n!)$$
$$ 0 + frac{a_2}{2} + 0 + frac{a_2}{4} + dots +frac{a_2}{n!} sim a_2 frac{gamma}{2} + frac{a_2}{2}ln(frac{n!}{2})$$
$$ 0 + 0 + frac{a_3}{3} + 0 + dots +frac{a_3}{n!} sim a_3 frac{gamma}{3} + frac{a_3}{3}ln(frac{n!}{3})$$
$$vdots $$
$+$
$-----------------------------------$
$$ underbrace{frac{b_1}{1}}_{a_1/1} + underbrace{frac{b_2}{2}}_{(a_1+ a_2)/2} + underbrace{frac{b_3}{3}}_{(a_1+ a_3)/3} + dots sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
In the above we define:
$$b_r = sum_{r|l} a_l text{ }forall text{ } 1 leq r leq n$$
$$b_r = sum_{(r-n)|l} a_l text{ }forall text{ } n+1 leq r leq 2n$$
$$b_r = sum_{(r-2n)|l} a_l text{ }forall text{ } 2n+1 leq r leq 3n$$
$$ vdots $$
$$b_r = sum_{(r-(n-1)!)|l} a_l text{ }forall text{ } n(n-1)!- n +1 leq r leq n!$$
Writing the above properly now:
$$ sum_{r=1}^n frac{b_r}{r} + sum_{r=n+1}^{2n} frac{b_{r-n}}{r} + dots+ sum_{r=n!-n+1}^{n!} frac{b_{r-n!+n }}{r} sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
Rearranging the L.H.S and R.H.S:
$$ sum_{r=1}^n sum_{k=0}^{(n-1)! -1 } b_r ( frac{1}{kn+ r}) sim (gamma+ ln n!) sum_{r=1}^n frac{a_r}{r} - sum_{r=1}^n frac{a_r}{r} ln(r) $$
Also note :
$$a_{l}=sum _{dmid l} mu left({frac {l}{d}}right)b_{d} implies frac{partial a_l}{ partial b_d} = mu left({frac {l}{d}}right) $$
where $mu$ is the mobius function. Let us now act $frac{partial }{partial b_d} $ on both sides where $d leq n$:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim (gamma+ ln n!) sum_{r=1}^n frac{mu(frac{r}{d})}{r} - sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
where $mu(lambda)= 0$ where $lambda$ is not an integer. Simplifying some of the terms:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
Let us take $d to 1$
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ 1}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r) $$
Now using some approximations:
$$ 1 + frac{1}{n} sum_{k=1}^{(n-1)! -1 } frac{1}{k} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
Using the first equation with a different $n$:
$$ 1 + frac{gamma + ln(n-1)!}{n} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
combinatorics number-theory proof-verification alternative-proof zeta-functions
asked Nov 20 '18 at 0:42
More Anonymous
34419
Background & Question
I realised I could do a different manipulation from the one I did over here Strange method to obtain strange number theoretic identities? However after making some calculations I seems to be getting the wrong answer by perhaps an extra term.
I got the expression:
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
But as $n to infty$ we know the $(frac{1}{zeta(k)})' to 0 $ when $k$ to $1$.
Is the correct answer below?
$$sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} $$
If so where did I go wrong in my proof? And what is the "normal" proof?
Proof
Let us write a relation of the Euler–Mascheroni constant for large $n$.
$$ 1+ frac{1}{2} + frac{1}{3} + frac{1}{4} + dots +frac{1}{n!} sim gamma + ln(n!)$$
Or, multiplying $1/2$ boths sides and let $n! to n!/2$
$$ 0 + frac{1}{2} + 0 + frac{1}{4} + dots +frac{1}{n!} sim frac{gamma}{2} + frac{1}{2}ln(frac{n!}{2})$$
Or, multiplying $1/3$ boths sides and let $n! to n!/3$
$$ 0 + 0 + frac{1}{3} + 0 + dots +frac{1}{n!} sim frac{gamma}{3} + frac{1}{3}ln(frac{n!}{3})$$
And so on $n$ times ... Now multiplying the $r$'th row with $a_r$ and adding vertically (whilst defining $b_r$):
$$ a_1+ frac{a_1}{2} + frac{a_1}{3} + frac{a_1}{4} + dots +frac{a_1}{n!} sim a_1 gamma + a_1 ln(n!)$$
$$ 0 + frac{a_2}{2} + 0 + frac{a_2}{4} + dots +frac{a_2}{n!} sim a_2 frac{gamma}{2} + frac{a_2}{2}ln(frac{n!}{2})$$
$$ 0 + 0 + frac{a_3}{3} + 0 + dots +frac{a_3}{n!} sim a_3 frac{gamma}{3} + frac{a_3}{3}ln(frac{n!}{3})$$
$$vdots $$
$+$
$-----------------------------------$
$$ underbrace{frac{b_1}{1}}_{a_1/1} + underbrace{frac{b_2}{2}}_{(a_1+ a_2)/2} + underbrace{frac{b_3}{3}}_{(a_1+ a_3)/3} + dots sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
In the above we define:
$$b_r = sum_{r|l} a_l text{ }forall text{ } 1 leq r leq n$$
$$b_r = sum_{(r-n)|l} a_l text{ }forall text{ } n+1 leq r leq 2n$$
$$b_r = sum_{(r-2n)|l} a_l text{ }forall text{ } 2n+1 leq r leq 3n$$
$$ vdots $$
$$b_r = sum_{(r-(n-1)!)|l} a_l text{ }forall text{ } n(n-1)!- n +1 leq r leq n!$$
Writing the above properly now:
$$ sum_{r=1}^n frac{b_r}{r} + sum_{r=n+1}^{2n} frac{b_{r-n}}{r} + dots+ sum_{r=n!-n+1}^{n!} frac{b_{r-n!+n }}{r} sim gamma sum_{r=1}^n frac{a_r}{r} + sum_{r=1}^n frac{a_r}{r} ln(frac{n!}{r}) $$
Rearranging the L.H.S and R.H.S:
$$ sum_{r=1}^n sum_{k=0}^{(n-1)! -1 } b_r ( frac{1}{kn+ r}) sim (gamma+ ln n!) sum_{r=1}^n frac{a_r}{r} - sum_{r=1}^n frac{a_r}{r} ln(r) $$
Also note :
$$a_{l}=sum _{dmid l} mu left({frac {l}{d}}right)b_{d} implies frac{partial a_l}{ partial b_d} = mu left({frac {l}{d}}right) $$
where $mu$ is the mobius function. Let us now act $frac{partial }{partial b_d} $ on both sides where $d leq n$:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim (gamma+ ln n!) sum_{r=1}^n frac{mu(frac{r}{d})}{r} - sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
where $mu(lambda)= 0$ where $lambda$ is not an integer. Simplifying some of the terms:
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ d}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(frac{r}{d})}{r} ln(r) $$
Let us take $d to 1$
$$ sum_{k=0}^{(n-1)! -1 } ( frac{1}{kn+ 1}) sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r) $$
Now using some approximations:
$$ 1 + frac{1}{n} sum_{k=1}^{(n-1)! -1 } frac{1}{k} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
Using the first equation with a different $n$:
$$ 1 + frac{gamma + ln(n-1)!}{n} sim frac{(gamma+ ln n!) }{n}- sum_{r=1}^n frac{mu(r)}{r} ln(r)$$
$$implies sum_{r=1}^n frac{mu(r)}{r} ln(r) sim frac{ln n}{n} -1 $$
combinatorics number-theory proof-verification alternative-proof zeta-functions
combinatorics number-theory proof-verification alternative-proof zeta-functions
asked Nov 20 '18 at 0:42
More Anonymous
34419
asked Nov 20 '18 at 0:42
More Anonymous
34419
edited Nov 20 '18 at 0:50
asked Nov 20 '18 at 0:42
More Anonymous
34419
asked Nov 20 '18 at 0:42
More Anonymous
34419
asked Nov 20 '18 at 0:42
More Anonymous
34419
34419
$sum_{n=1}^infty frac{mu(n)}{n} (-log n)^k$ converges for every $k$ (prime number theorem) and the rate of convergence encodes the zero-free region of $zeta(s)$ (thus the Riemann hypothesis). In particular $sum_{n=1}^N frac{mu(n)}{n} (-log n)^k= frac{d^k}{ds^k} frac{1}{zeta(s)}|_{s=1}+ O(N^{-1+sigma+epsilon})$ iff $zeta(s)$ has no zeros for $Re(s) > sigma$.
– reuns
Nov 20 '18 at 1:04
@reuns So this is a highly non-trivial statement I'm trying to make?
– More Anonymous
Nov 20 '18 at 1:07
I'll be editing this post (I caught a foolish error)
– More Anonymous
Nov 20 '18 at 2:28
add a comment |
$sum_{n=1}^infty frac{mu(n)}{n} (-log n)^k$ converges for every $k$ (prime number theorem) and the rate of convergence encodes the zero-free region of $zeta(s)$ (thus the Riemann hypothesis). In particular $sum_{n=1}^N frac{mu(n)}{n} (-log n)^k= frac{d^k}{ds^k} frac{1}{zeta(s)}|_{s=1}+ O(N^{-1+sigma+epsilon})$ iff $zeta(s)$ has no zeros for $Re(s) > sigma$.
– reuns
Nov 20 '18 at 1:04
@reuns So this is a highly non-trivial statement I'm trying to make?
– More Anonymous
Nov 20 '18 at 1:07
I'll be editing this post (I caught a foolish error)
– More Anonymous
Nov 20 '18 at 2:28
$sum_{n=1}^infty frac{mu(n)}{n} (-log n)^k$ converges for every $k$ (prime number theorem) and the rate of convergence encodes the zero-free region of $zeta(s)$ (thus the Riemann hypothesis). In particular $sum_{n=1}^N frac{mu(n)}{n} (-log n)^k= frac{d^k}{ds^k} frac{1}{zeta(s)}|_{s=1}+ O(N^{-1+sigma+epsilon})$ iff $zeta(s)$ has no zeros for $Re(s) > sigma$.
– reuns
Nov 20 '18 at 1:04
$sum_{n=1}^infty frac{mu(n)}{n} (-log n)^k$ converges for every $k$ (prime number theorem) and the rate of convergence encodes the zero-free region of $zeta(s)$ (thus the Riemann hypothesis). In particular $sum_{n=1}^N frac{mu(n)}{n} (-log n)^k= frac{d^k}{ds^k} frac{1}{zeta(s)}|_{s=1}+ O(N^{-1+sigma+epsilon})$ iff $zeta(s)$ has no zeros for $Re(s) > sigma$.
– reuns
Nov 20 '18 at 1:04
@reuns So this is a highly non-trivial statement I'm trying to make?
– More Anonymous
Nov 20 '18 at 1:07
@reuns So this is a highly non-trivial statement I'm trying to make?
– More Anonymous
Nov 20 '18 at 1:07
I'll be editing this post (I caught a foolish error)
– More Anonymous
Nov 20 '18 at 2:28
I'll be editing this post (I caught a foolish error)
– More Anonymous
Nov 20 '18 at 2:28
add a comment |
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$sum_{n=1}^infty frac{mu(n)}{n} (-log n)^k$ converges for every $k$ (prime number theorem) and the rate of convergence encodes the zero-free region of $zeta(s)$ (thus the Riemann hypothesis). In particular $sum_{n=1}^N frac{mu(n)}{n} (-log n)^k= frac{d^k}{ds^k} frac{1}{zeta(s)}|_{s=1}+ O(N^{-1+sigma+epsilon})$ iff $zeta(s)$ has no zeros for $Re(s) > sigma$.
– reuns
Nov 20 '18 at 1:04
@reuns So this is a highly non-trivial statement I'm trying to make?
– More Anonymous
Nov 20 '18 at 1:07
I'll be editing this post (I caught a foolish error)
– More Anonymous
Nov 20 '18 at 2:28