Servois' 1817 "Memoir on Quadratures" – Bernoulli Numbers in Servois' Memoir

The Bernoulli Numbers appear in Servois' memoir in a curious way. Because

\[\sum y = (E-1)^{-1} y \quad \mbox{and} \quad Ey = e^d y\]

we must have

\[ (e^d - 1)\sum y = y.\,\,\,\,\,\,\,\,\,\,\,\mbox{(VII)}\]

Servois assumed that the operator \(\sum\) has the form \(Ad^{-1} + Bd^0 + C d^1 + D d^2 + \cdots\) and, using the series expansion of \(e^d\), solved the operator version of equation (VII):

\[\left(d + \frac{d^2}{2!} + \frac{d^3}{3!} + \cdots\right) \left(Ad^{-1} + B + C d + Dd^2 + \cdots \right) = 1.\]

Expanding the left hand side, one term of the first series at a time, we get

\[\begin{array}{rllllllllll} A &+& Bd        &+& Cd^2        &+& Dd^3         &+& E d^4        &+& \cdots \\ \\  &+& \frac{A}{2} d &+&\frac{B}{2} d^2 &+& \frac{C}{2} d^3  &+& \frac{D}{2} d^4 &+& \cdots \\ \\  &&            &+& \frac{A}{6} d^2 &+& \frac{B}{6} d^3  &+& \frac{C}{6} d^4  &+& \cdots \\ \\  &&            &&             &+& \frac{A}{24} d^3 &+& \frac{B}{24} d^4 &+& \cdots \\ \\  &&       &&          &&          &+& \cdots        &+& \cdots. \end{array}\]

Now sum the columns: the first column sums to \(1\) and the others to \(0\). This gives

\[A=1, \quad B=-\frac{1}{2}, \quad C=\frac{\frac{1}{6}}{2!}, \quad D=0, \quad E=\frac{-\frac{1}{30}}{4!}, \ldots .\,\,\,\,\,\,\,\,\,\,\,\mbox{(VIII)}\]

That is, the numbers \(A, B, C, D, E, \ldots\) are equal to

\[\frac{b_k}{k!}\]

for \(k=0, 1, 2, \ldots\), with the exception that \(B=-b_1\). This is why the Bernoulli numbers appear in Servois' equation (6).