Mixing
Harmonic series for the logarithm of integer ratios
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
A paper has the following bound (without explanation) :
For $nin mathbbN$ and $N$ large,
$$sum_m leq N, m neq n frac1 leq C N log N$$
where $C$ is a universal constant.
How do you prove it ?
If $m,n$ are prime numbers, what would be the upper bound ?
real-analysis number-theory analytic-number-theory
asked Jul 16 at 16:20
Frédéric Ouimet
366
add a comment |Â
up vote
3
down vote
favorite
A paper has the following bound (without explanation) :
For $nin mathbbN$ and $N$ large,
$$sum_m leq N, m neq n frac1 leq C N log N$$
where $C$ is a universal constant.
How do you prove it ?
If $m,n$ are prime numbers, what would be the upper bound ?
real-analysis number-theory analytic-number-theory
asked Jul 16 at 16:20
Frédéric Ouimet
366
Check the references of the paper probably for the bound. I would imagine that the proof or explanation of the bound would be there.
– Eleven-Eleven
Jul 16 at 16:27
It's not. It's supposed to be a relatively easy analytic number theory argument, but it's not my field.
– Frédéric Ouimet
Jul 16 at 16:35
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
A paper has the following bound (without explanation) :
For $nin mathbbN$ and $N$ large,
$$sum_m leq N, m neq n frac1 leq C N log N$$
where $C$ is a universal constant.
How do you prove it ?
If $m,n$ are prime numbers, what would be the upper bound ?
real-analysis number-theory analytic-number-theory
asked Jul 16 at 16:20
Frédéric Ouimet
366
A paper has the following bound (without explanation) :
For $nin mathbbN$ and $N$ large,
$$sum_m leq N, m neq n frac1 leq C N log N$$
where $C$ is a universal constant.
How do you prove it ?
If $m,n$ are prime numbers, what would be the upper bound ?
real-analysis number-theory analytic-number-theory
asked Jul 16 at 16:20
Frédéric Ouimet
366
asked Jul 16 at 16:20
Frédéric Ouimet
366
asked Jul 16 at 16:20
Frédéric Ouimet
366
asked Jul 16 at 16:20
Frédéric Ouimet
366
366
Check the references of the paper probably for the bound. I would imagine that the proof or explanation of the bound would be there.
– Eleven-Eleven
Jul 16 at 16:27
It's not. It's supposed to be a relatively easy analytic number theory argument, but it's not my field.
– Frédéric Ouimet
Jul 16 at 16:35
add a comment |Â
Check the references of the paper probably for the bound. I would imagine that the proof or explanation of the bound would be there.
– Eleven-Eleven
Jul 16 at 16:27
It's not. It's supposed to be a relatively easy analytic number theory argument, but it's not my field.
– Frédéric Ouimet
Jul 16 at 16:35
Check the references of the paper probably for the bound. I would imagine that the proof or explanation of the bound would be there.
– Eleven-Eleven
Jul 16 at 16:27
Check the references of the paper probably for the bound. I would imagine that the proof or explanation of the bound would be there.
– Eleven-Eleven
Jul 16 at 16:27
It's not. It's supposed to be a relatively easy analytic number theory argument, but it's not my field.
– Frédéric Ouimet
Jul 16 at 16:35
It's not. It's supposed to be a relatively easy analytic number theory argument, but it's not my field.
– Frédéric Ouimet
Jul 16 at 16:35
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You can do this with some quick and dirty bounds:
Let's break the sum up into three pieces:
$m leq n/2$
$m in [n/2,2n]$
$m > 2n$
For $m < n/2$ and $m > 2n$, each summand is bounded above by $1/log(2)$, implying that the sums over that region are bounded above by $2N/log(2)$. We just have to deal with the middle part.
For $c > 0$ sufficiently small, we have a bound of $|log(1 + x)| > c |x|$ for all $x in [1/2,2]$. Therefore, for $m in [n/2,2n]$, write $m = n + a$ and we have bounds $$frac1 = frac1 leq frac1a = fracn,.$$
Summing over $a in [-n/2,n] setminus0$ gives two harmonic series, each of which is bounded above by $Clog(n)$. Finally, bounding $n$ by $N$ completes the proof.
EDIT: Also worth mentioning, nothing special happens with prime numbers; there's not really any arithmetic structure going on here.
answered Jul 16 at 18:25
Marcus M
8,1731847
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You can do this with some quick and dirty bounds:
Let's break the sum up into three pieces:
$m leq n/2$
$m in [n/2,2n]$
$m > 2n$
For $m < n/2$ and $m > 2n$, each summand is bounded above by $1/log(2)$, implying that the sums over that region are bounded above by $2N/log(2)$. We just have to deal with the middle part.
For $c > 0$ sufficiently small, we have a bound of $|log(1 + x)| > c |x|$ for all $x in [1/2,2]$. Therefore, for $m in [n/2,2n]$, write $m = n + a$ and we have bounds $$frac1 = frac1 leq frac1a = fracn,.$$
Summing over $a in [-n/2,n] setminus0$ gives two harmonic series, each of which is bounded above by $Clog(n)$. Finally, bounding $n$ by $N$ completes the proof.
EDIT: Also worth mentioning, nothing special happens with prime numbers; there's not really any arithmetic structure going on here.
answered Jul 16 at 18:25
Marcus M
8,1731847
add a comment |Â
up vote
3
down vote
accepted
You can do this with some quick and dirty bounds:
Let's break the sum up into three pieces:
$m leq n/2$
$m in [n/2,2n]$
$m > 2n$
For $m < n/2$ and $m > 2n$, each summand is bounded above by $1/log(2)$, implying that the sums over that region are bounded above by $2N/log(2)$. We just have to deal with the middle part.
For $c > 0$ sufficiently small, we have a bound of $|log(1 + x)| > c |x|$ for all $x in [1/2,2]$. Therefore, for $m in [n/2,2n]$, write $m = n + a$ and we have bounds $$frac1 = frac1 leq frac1a = fracn,.$$
Summing over $a in [-n/2,n] setminus0$ gives two harmonic series, each of which is bounded above by $Clog(n)$. Finally, bounding $n$ by $N$ completes the proof.
EDIT: Also worth mentioning, nothing special happens with prime numbers; there's not really any arithmetic structure going on here.
answered Jul 16 at 18:25
Marcus M
8,1731847
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You can do this with some quick and dirty bounds:
Let's break the sum up into three pieces:
$m leq n/2$
$m in [n/2,2n]$
$m > 2n$
For $m < n/2$ and $m > 2n$, each summand is bounded above by $1/log(2)$, implying that the sums over that region are bounded above by $2N/log(2)$. We just have to deal with the middle part.
For $c > 0$ sufficiently small, we have a bound of $|log(1 + x)| > c |x|$ for all $x in [1/2,2]$. Therefore, for $m in [n/2,2n]$, write $m = n + a$ and we have bounds $$frac1 = frac1 leq frac1a = fracn,.$$
Summing over $a in [-n/2,n] setminus0$ gives two harmonic series, each of which is bounded above by $Clog(n)$. Finally, bounding $n$ by $N$ completes the proof.
EDIT: Also worth mentioning, nothing special happens with prime numbers; there's not really any arithmetic structure going on here.
answered Jul 16 at 18:25
Marcus M
8,1731847
You can do this with some quick and dirty bounds:
Let's break the sum up into three pieces:
$m leq n/2$
$m in [n/2,2n]$
$m > 2n$
For $m < n/2$ and $m > 2n$, each summand is bounded above by $1/log(2)$, implying that the sums over that region are bounded above by $2N/log(2)$. We just have to deal with the middle part.
For $c > 0$ sufficiently small, we have a bound of $|log(1 + x)| > c |x|$ for all $x in [1/2,2]$. Therefore, for $m in [n/2,2n]$, write $m = n + a$ and we have bounds $$frac1 = frac1 leq frac1a = fracn,.$$
Summing over $a in [-n/2,n] setminus0$ gives two harmonic series, each of which is bounded above by $Clog(n)$. Finally, bounding $n$ by $N$ completes the proof.
EDIT: Also worth mentioning, nothing special happens with prime numbers; there's not really any arithmetic structure going on here.
answered Jul 16 at 18:25
Marcus M
8,1731847
answered Jul 16 at 18:25
Marcus M
8,1731847
answered Jul 16 at 18:25
Marcus M
8,1731847
answered Jul 16 at 18:25
Marcus M
8,1731847
8,1731847
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853562%2fharmonic-series-for-the-logarithm-of-integer-ratios%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Check the references of the paper probably for the bound. I would imagine that the proof or explanation of the bound would be there.
– Eleven-Eleven
Jul 16 at 16:27
It's not. It's supposed to be a relatively easy analytic number theory argument, but it's not my field.
– Frédéric Ouimet
Jul 16 at 16:35