Introductio in analysin infinitorum

CONTINUIS. 307 CAP. XVIII. $b = A;$ $\alpha = 1$ $c = B - 1;$ $\beta = A$ $d = C - 1;$ $\gamma = B$ $e = D - 1;$ $\delta = C$ $f = E - 1;$ $\epsilon = D$ &c., unde consequenter fiet $x = \frac{1}{A+\frac{A}{B-1+\frac{B}{C-1+\frac{C}{D-1+\frac{D}{E-1+\&c.}}}}}$ EXEMPLUM I. Quoniam, posito $e$ numero cujus Logarithmus est $= 1$, supra invenimus esse $\frac{1}{e} = 1 - \frac{1}{1} + \frac{1}{1 \cdot 2} - \frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} - \&c.,$ feu $1 - \frac{1}{e} = \frac{1}{1} - \frac{1}{1 \cdot 2} + \frac{1}{1 \cdot 2 \cdot 3} - \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \&c.,$ hæc [TRUNCATED]