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Summing Bernoulli numbers
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Consider the Bernoulli numbers denoted by $B_n$, which are rational numbers.
It is known that the harmonic numbers $H_n=sum_k=1^nfrac1k$ are not integers once $n>1$.
I am curious about the following:
Question: If $n>0$, will $sum_k=0^nB_k$ ever be an integer?
nt.number-theory real-analysis
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
6
down vote
favorite
Consider the Bernoulli numbers denoted by $B_n$, which are rational numbers.
It is known that the harmonic numbers $H_n=sum_k=1^nfrac1k$ are not integers once $n>1$.
I am curious about the following:
Question: If $n>0$, will $sum_k=0^nB_k$ ever be an integer?
nt.number-theory real-analysis
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Consider the Bernoulli numbers denoted by $B_n$, which are rational numbers.
It is known that the harmonic numbers $H_n=sum_k=1^nfrac1k$ are not integers once $n>1$.
I am curious about the following:
Question: If $n>0$, will $sum_k=0^nB_k$ ever be an integer?
nt.number-theory real-analysis
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
Consider the Bernoulli numbers denoted by $B_n$, which are rational numbers.
It is known that the harmonic numbers $H_n=sum_k=1^nfrac1k$ are not integers once $n>1$.
I am curious about the following:
Question: If $n>0$, will $sum_k=0^nB_k$ ever be an integer?
nt.number-theory real-analysis
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
asked Sep 6 at 18:23
T. Amdeberhan
15.7k225119
15.7k225119
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
12
down vote
accepted
It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that
$$B_n+sum_nfrac1pin mathbb Z$$
[1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen.
Astr. Nachr., 17:351–352, 1840
[2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen
betreffend. J. Reine Angew. Math., 21:372–374, 1840
Using this we see that the integrality of $sum_k=0^n B_k$ is equivalent to the integrality of
$$sum_k=2^nleft(sum_pin mathbb P , , , p-1frac1pright)=sum_pin mathbb Pfraclfloorfracnp-1rfloorp$$
If $q$ is the largest prime $le n$ we have $lfloor fracnq-1rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
add a comment |Â
up vote
8
down vote
It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_p-1$ and not of other summands in your sum, by von Staudt - Clausen Theorem.
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that
$$B_n+sum_nfrac1pin mathbb Z$$
[1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen.
Astr. Nachr., 17:351–352, 1840
[2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen
betreffend. J. Reine Angew. Math., 21:372–374, 1840
Using this we see that the integrality of $sum_k=0^n B_k$ is equivalent to the integrality of
$$sum_k=2^nleft(sum_pin mathbb P , , , p-1frac1pright)=sum_pin mathbb Pfraclfloorfracnp-1rfloorp$$
If $q$ is the largest prime $le n$ we have $lfloor fracnq-1rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
add a comment |Â
up vote
12
down vote
accepted
It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that
$$B_n+sum_nfrac1pin mathbb Z$$
[1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen.
Astr. Nachr., 17:351–352, 1840
[2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen
betreffend. J. Reine Angew. Math., 21:372–374, 1840
Using this we see that the integrality of $sum_k=0^n B_k$ is equivalent to the integrality of
$$sum_k=2^nleft(sum_pin mathbb P , , , p-1frac1pright)=sum_pin mathbb Pfraclfloorfracnp-1rfloorp$$
If $q$ is the largest prime $le n$ we have $lfloor fracnq-1rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that
$$B_n+sum_nfrac1pin mathbb Z$$
[1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen.
Astr. Nachr., 17:351–352, 1840
[2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen
betreffend. J. Reine Angew. Math., 21:372–374, 1840
Using this we see that the integrality of $sum_k=0^n B_k$ is equivalent to the integrality of
$$sum_k=2^nleft(sum_pin mathbb P , , , p-1frac1pright)=sum_pin mathbb Pfraclfloorfracnp-1rfloorp$$
If $q$ is the largest prime $le n$ we have $lfloor fracnq-1rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that
$$B_n+sum_nfrac1pin mathbb Z$$
[1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen.
Astr. Nachr., 17:351–352, 1840
[2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen
betreffend. J. Reine Angew. Math., 21:372–374, 1840
Using this we see that the integrality of $sum_k=0^n B_k$ is equivalent to the integrality of
$$sum_k=2^nleft(sum_pin mathbb P , , , p-1frac1pright)=sum_pin mathbb Pfraclfloorfracnp-1rfloorp$$
If $q$ is the largest prime $le n$ we have $lfloor fracnq-1rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
answered Sep 6 at 18:42
Gjergji Zaimi
58.7k3153291
58.7k3153291
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
add a comment |Â
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
Thank you, Gjergji.
– T. Amdeberhan
Sep 7 at 17:17
add a comment |Â
up vote
8
down vote
It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_p-1$ and not of other summands in your sum, by von Staudt - Clausen Theorem.
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
add a comment |Â
up vote
8
down vote
It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_p-1$ and not of other summands in your sum, by von Staudt - Clausen Theorem.
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
add a comment |Â
up vote
8
down vote
up vote
8
down vote
It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_p-1$ and not of other summands in your sum, by von Staudt - Clausen Theorem.
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_p-1$ and not of other summands in your sum, by von Staudt - Clausen Theorem.
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
answered Sep 6 at 18:41
Fedor Petrov
44.6k5107211
44.6k5107211
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
add a comment |Â
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
Thank you, Fedor.
– T. Amdeberhan
Sep 7 at 13:20
add a comment |Â
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