Floor of Riemann zeta functionCertain functional equations for the Riemann Zeta function? Is it true that the sum of a specific floor function of a prime = 1?Mellin inverse of the Hadamard product rep. of the Riemann zeta function?Help with an irregular integralAnalytic continuation of the Dirichlet generating series of the multiplicative partition functionAre the twin primes the only positive double zeros of this real function?expressing $log(left lfloor x right rfloor!)$ in terms of zeta-zerosSome identities with the Riemann-Hurwitz zeta functionA question involving e, floor, and all x > 0About the logarithmic derivative of the Riemann zeta function
Floor of Riemann zeta function
Certain functional equations for the Riemann Zeta function? Is it true that the sum of a specific floor function of a prime = 1?Mellin inverse of the Hadamard product rep. of the Riemann zeta function?Help with an irregular integralAnalytic continuation of the Dirichlet generating series of the multiplicative partition functionAre the twin primes the only positive double zeros of this real function?expressing $log(left lfloor x right rfloor!)$ in terms of zeta-zerosSome identities with the Riemann-Hurwitz zeta functionA question involving e, floor, and all x > 0About the logarithmic derivative of the Riemann zeta function
$begingroup$
How to show that $$leftlfloorzetaleft(1+frac1nright)rightrfloor=n$$ for every positive integer $n$?
nt.number-theory cv.complex-variables analytic-number-theory
edited 2 hours ago
GH from MO
59.5k5150228
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
$endgroup$
add a comment |
$begingroup$
How to show that $$leftlfloorzetaleft(1+frac1nright)rightrfloor=n$$ for every positive integer $n$?
nt.number-theory cv.complex-variables analytic-number-theory
edited 2 hours ago
GH from MO
59.5k5150228
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
$endgroup$
add a comment |
$begingroup$
How to show that $$leftlfloorzetaleft(1+frac1nright)rightrfloor=n$$ for every positive integer $n$?
nt.number-theory cv.complex-variables analytic-number-theory
edited 2 hours ago
GH from MO
59.5k5150228
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
$endgroup$
How to show that $$leftlfloorzetaleft(1+frac1nright)rightrfloor=n$$ for every positive integer $n$?
nt.number-theory cv.complex-variables analytic-number-theory
nt.number-theory cv.complex-variables analytic-number-theory
edited 2 hours ago
GH from MO
59.5k5150228
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
edited 2 hours ago
GH from MO
59.5k5150228
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
edited 2 hours ago
GH from MO
59.5k5150228
edited 2 hours ago
GH from MO
59.5k5150228
edited 2 hours ago
GH from MO
59.5k5150228
59.5k5150228
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
asked 2 hours ago
Gianni del FioreGianni del Fiore
855
855
add a comment |
add a comment |
1 Answer
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It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I)
$$frac1sigma-1<zeta(sigma)<fracsigmasigma-1,qquad sigmain(0,1)cup(1,infty).$$
In particular, taking $sigma=1+frac1n$, we get
$$n<zetaleft(1+frac1nright)<n+1.$$
This is slightly stronger than your claim. Better bounds can be obtained from the Laurent series expansion of $zeta(s)$ around $s=1$.
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
$endgroup$
1
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
add a comment |
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1 Answer
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1 Answer
1
active
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$begingroup$
It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I)
$$frac1sigma-1<zeta(sigma)<fracsigmasigma-1,qquad sigmain(0,1)cup(1,infty).$$
In particular, taking $sigma=1+frac1n$, we get
$$n<zetaleft(1+frac1nright)<n+1.$$
This is slightly stronger than your claim. Better bounds can be obtained from the Laurent series expansion of $zeta(s)$ around $s=1$.
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
$endgroup$
1
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
add a comment |
$begingroup$
It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I)
$$frac1sigma-1<zeta(sigma)<fracsigmasigma-1,qquad sigmain(0,1)cup(1,infty).$$
In particular, taking $sigma=1+frac1n$, we get
$$n<zetaleft(1+frac1nright)<n+1.$$
This is slightly stronger than your claim. Better bounds can be obtained from the Laurent series expansion of $zeta(s)$ around $s=1$.
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
$endgroup$
1
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
add a comment |
$begingroup$
It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I)
$$frac1sigma-1<zeta(sigma)<fracsigmasigma-1,qquad sigmain(0,1)cup(1,infty).$$
In particular, taking $sigma=1+frac1n$, we get
$$n<zetaleft(1+frac1nright)<n+1.$$
This is slightly stronger than your claim. Better bounds can be obtained from the Laurent series expansion of $zeta(s)$ around $s=1$.
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
$endgroup$
It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I)
$$frac1sigma-1<zeta(sigma)<fracsigmasigma-1,qquad sigmain(0,1)cup(1,infty).$$
In particular, taking $sigma=1+frac1n$, we get
$$n<zetaleft(1+frac1nright)<n+1.$$
This is slightly stronger than your claim. Better bounds can be obtained from the Laurent series expansion of $zeta(s)$ around $s=1$.
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
answered 2 hours ago
GH from MOGH from MO
59.5k5150228
59.5k5150228
1
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
add a comment |
1
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
1
1
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
$begingroup$
thank you very much for your answer
$endgroup$
– Gianni del Fiore
2 hours ago
add a comment |
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