Consider the [Bernoulli numbers](http://mathworld.wolfram.com/BernoulliNumber.html) denoted by $B\_n$, which are rational numbers.


It is known that the harmonic numbers $H\_n=\sum\_{k=1}^n\frac1k$ 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?
> 
> 
> 



Summing Bernoulli numbers
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
$$B\_n+\sum\_{p\in \mathbb{P}\, ,\, p-1|n}\frac{1}{p}\in \mathbb Z$$



> 
> [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
$$\sum\_{k=2}^n\left(\sum\_{p\in \mathbb P \, , \, p-1|k}\frac{1}{p}\right)=\sum\_{p\in \mathbb P}\frac{\lfloor\frac{n}{p-1}\rfloor}{p}$$
If $q$ is the largest prime $\le n$ we have $\lfloor \frac{n}{q-1}\rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.


