Summing Bernoulli numbers

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?

2 Answers
2

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
$$B_n+sum_pin mathbbP, ,, p-1frac1pin mathbb Z$$

[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.

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.

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