Introductio in analysin infinitorum

32 DE TRANSFORMATIONE LIB. I. posito $1-z=0$ seu $z=1$; ex quo prodit $A=\frac{1}{2}$. Jam ponatur $T = \frac{M-AS}{1-z}$, erit $T = \frac{\frac{1}{2}zz-\frac{1}{2}}{1-z} = -\frac{1}{2} - \frac{1}{2}z$; unde oritur $B = -\frac{1}{2}$, posito $z=1$, ita ut fit $B = -\frac{1}{2}$. Ponatur porro $V = \frac{T-BS}{1-z} = \frac{T+\frac{1}{2}S}{1-z}$; erit $V = \frac{-\frac{1}{2}z+\frac{1}{2}zz}{1-z} = -\frac{1}{2}z$; unde fit $C = \frac{V}{S} = \frac{-\frac{1}{2}z}{1+zz}$ posito $z=1$, ita ut fit $C = -\frac{1}{4}$. Quo circa fractiones partiales ex denominatoris Factore $(1-z)^3$ ortae erunt $\frac{1}{2(1-z)^3} - \frac{1}{2(1-z)^2} - \frac{1}{4(1-z)}$. 45. Si Functionis [TRUNCATED]