Question

Addendum. Since I had already TeXed Witt's proof for my Notes, it is easy to reproduce it here :

Theorem (von Staudt--Clausen, 1840)
Let $k>0$ be an even integer, and let $l$ run through the primes. Then the(1)\quad\quad\quad W_k=B_k+\sum_{l-1|k}{1\over l}is always an integer. For example, $\displaystyleW_{12}=B_{12}+{1\over 2}+{1\over3}+{1\over5}+{1\over7}+{1\over 13}=1$.

[The British analyst Hardy says in his Twelve lectures (p. 11) that this theorem was rediscovered by Ramanujan ``at a time of his life when he had hardly formed any definite concept of proof''.]

Proof (Witt) : The idea is to show that $W_k$ is a $p$-adic integerfor every prime $p$. More precisely, we show that $B_k+p^{-1}$ (resp. $B_k$)is a $p$-adic integer if $p-1|k$ (resp. if not).

For an integer $n>0$, let $S_k(n)=0^k+1^k+2^k+\cdots+(n-1)^k$. Comparing thecoefficients on the two sides of1+e^T+e^{2T}+\cdots+e^{(n-1)T}={e^{nT}-1\over T}{T\over e^T-1},we get $\displaystyle S_k(n) =\sum_{m\in[0,k]}{k\choose m}{B_m\over k+1-m}n^{k+1-m}$. To recover $B_k$ from the $S_k(n)$, it is tempting totake the limit $\displaystyle\lim_{n\to0}S_k(n)/n$, which doesn't make sensein the archimedean world. If, however, we make $n$ run through the powers$p^s$ of a fixed prime $p$, then, $p$-adically, $p^s\to0$ as $s\to+\infty$,Let us compare $S_k(p^{s+1})/p^{s+1}$ with $S_k(p^s)/p^s$. Every$j\in[0,p^{s+1}[$ can be uniquely written as $j=up^s+v$, where $u\in[0,p[$ and$v\in[0,p^s[$. Now,% &\equiv p\left(\sum_{v\in[0,p^s[}v^k\right)% +kp^s\left(\sum_{u\in[0,p[}u\sum_{v\in[0,p^s[}v^{k-1}\right)% \pmod{p^{2s}}\cr
&\equiv p\left(\sum_v v^k\right) +kp^s\left(\sum_u u\sum_v v^{k-1}\right)\pmod{p^{2s}}\cr by the binomial theorem. As $\sum_{v}v^k=S_k(p^s)$ and$2\sum_uu=p(p-1)\equiv0\pmod p$, we getS_k(p^{s+1})\equiv pS_k(p^s)\pmod{p^{s+1}},where, for $p=2$, the fact that $k$ is even has been used. Dividingthroughout by $p^{s+1}$, this can be expressed by saying that{S_k(p^{s+1})\over p^{s+1}}-{S_k(p^s)\over p^s}\in{\mathbf Z}_{(p)}is a $p$-adic integer, and therefore{S_k(p^r)\over p^r}-{S_k(p^s)\over p^s}\in Z_{(p)}for any two integers $r>0$, $s>0$, since ${\mathbf Z}_{(p)}$ is a subring of $\mathbf Q$.Fixing $s=1$ and letting $r\to+\infty$, we see that $B_k-S_k(p)/p\in{\mathbf Z}_{(p)}$,in view of $(2)$. We need a

Lemma.$S_k(p)=\sum_{j\in[1,p[}j^k$ is $\equiv-1\pmod p$ if $p-1|k$ and $\equiv0\pmod p$ otherwise.

This is clear if $p-1|k$. Suppose not, and let $g$ be a generator of$({\mathbf Z}/p{\mathbf Z})^\times$. We have $g^k-1\not\equiv0$, whereas\equiv (g^k-1)\left(\sum_{t\in[0,p-1[}g^{tk}\right)\equiv g^{(p-1)k}-1\equiv0.It follows that $B_k+p^{-1}\in{\mathbf Z}_{(p)}$ if $p-1|k$ and $B_k\in{\mathbf Z}_{(p)}$otherwise. In either case, the number $W_k$ (1), which can be written as(B_k+p^{-1})+\sum_{l\neq p}l^{-1}&\hbox{if }p-1|k\cr(B_k)+\sum_l l^{-1}&\hbox{otherwise},\cr}$$ (where $l$ runs through the primes for which $l-1|k$) turns out to be a$p$-adic integer for every prime $p$. Hence $W_k\in{\mathbf Z}$, as claimed.