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).