Editing Bernoulli number (section)

=== Why do the odd Bernoulli numbers vanish? ===
The sum

:<math>\varphi_k(n) = \sum_{i=0}^n i^k - \frac{n^k} 2</math>

can be evaluated for negative values of the index {{math|''n''}}. Doing so will show that it is an [[odd function]] for even values of {{math|''k''}}, which implies that the  sum has only terms of odd index. This and the formula for the Bernoulli sum imply that {{math|''B''<sub>2''k'' + 1 − ''m''</sub>}} is 0 for {{math|''m''}} even and {{math|2''k'' + 1 − ''m'' > 1}}; and that the term for {{math|''B''<sub>1</sub>}} is cancelled by the subtraction.  The von Staudt–Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid for ''n'' > 1).

From the von Staudt–Clausen theorem it is known that for odd {{math|''n'' > 1}} the number {{math|2''B''<sub>''n''</sub>}} is an integer. This seems trivial if one knows beforehand that the integer in question is zero. However, by applying Worpitzky's representation one gets

: <math> 2B_n =\sum_{m=0}^n (-1)^m \frac{2}{m+1}m! \left\{{n+1\atop m+1} \right\} = 0\quad(n>1 \text{ is odd})</math>

as a ''sum of integers'', which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. Let {{math|''S''<sub>''n'',''m''</sub>}} be the number of surjective maps from {{math|1={1, 2, ..., ''n''}}} to {{math|1={1, 2, ..., ''m''}}}, then {{math|''S''<sub>''n'',''m''</sub> {{=}} ''m''!<big><big>{</big></big>{{su|p=''n''|b=''m''|a=c}}<big><big>}</big></big>}}. The last equation can only hold if

: <math> \sum_{\text{odd }m=1}^{n-1} \frac 2 {m^2}S_{n,m}=\sum_{\text{even } m=2}^n \frac{2}{m^2} S_{n,m} \quad (n>2 \text{ is even}). </math>

This equation can be proved by induction. The first two examples of this equation are

:{{math|1=''n'' = 4: 2 + 8 = 7 + 3}},
:{{math|1=''n'' = 6: 2 + 120 + 144 = 31 + 195 + 40}}.

Thus the Bernoulli numbers vanish at odd index because some non-obvious combinatorial identities are embodied in the Bernoulli numbers.

=== A restatement of the Riemann hypothesis ===
The connection between the Bernoulli numbers and the Riemann zeta function is strong enough to provide an alternate formulation of the [[Riemann hypothesis]] (RH) which uses only the Bernoulli numbers. In fact [[Marcel Riesz]] proved that the RH is equivalent to the following assertion:{{r|Riesz1916}}

:For every {{math|''ε'' > {{sfrac|1|4}}}} there exists a constant {{math|''C''<sub>''ε''</sub> > 0}} (depending on {{math|''ε''}}) such that {{math|{{abs|''R''(''x'')}} < ''C''<sub>''ε''</sub>''x''<sup>''ε''</sup>}} as {{math|''x'' → ∞}}.

Here {{math|''R''(''x'')}} is the [[Riesz function]]

: <math> R(x) = 2 \sum_{k=1}^\infty
\frac{k^{\overline{k}} x^{k}}{(2\pi)^{2k}\left(\frac{B_{2k}}{2k}\right)}
= 2\sum_{k=1}^\infty \frac{k^{\overline{k}}x^k}{(2\pi)^{2k}\beta_{2k}}. </math>

{{math|''n''<sup>{{overline|''k''}}</sup>}} denotes the [[Pochhammer symbol#Alternate notations|rising factorial power]] in the notation of [[D. E. Knuth]]. The numbers {{math|''β''<sub>''n''</sub> {{=}} {{sfrac|''B''<sub>''n''</sub>|''n''}}}} occur frequently in the study of the zeta function and are significant  because {{math|''β''<sub>''n''</sub>}} is a {{math|''p''}}-integer for primes {{math|''p''}} where {{math|''p'' − 1}} does not divide {{math|''n''}}.  The {{math|''β''<sub>''n''</sub>}} are called ''divided Bernoulli numbers''.