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

Author(s):

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).

Robert E. Bradley (Adelphi University) and Salvatore J. Petrilli, Jr. (Adelphi University), "Servois' 1817 "Memoir on Quadratures" – Bernoulli Numbers in Servois' Memoir," Convergence (May 2019)