Introductio in analysin infinitorum

RECURRENTIBUS. 177 $C=C; D=D; \&c.$ Hoc autem dubium facile tolletur, si perpendamus hanc æqualitatem subsistere debere quemcunque valorem obtineat variabilis $z$. Sit igitur $z=0$, atque manifestum est fore $A=A$. His ergo terminis æqualibus utrinque sublatis, ac reliqua æquatione per $z$ divisa, habebitur $B + Cz + Dz^2 + \&c. = B + Cz + Dz^2 + \&c.;$ unde sequitur fore $B=B$: simili autem modo ostendetur esse $C=C; D=D$, & ita porro in infinitum. 215. Contemplemur ergo Series, quæ ex fractionibus partialibus, in quas fractio quæpiam proposita resolvitur, oriuntur. Ac primo quidem patet fractionem $\frac{A}{1-pz}$ dare Seriem $A + Apz + Ap^2z^2 + Ap^3z^3 + \&c.$, cujus terminus generalis est $Ap^nz^n$; hæc enim expressio vocari solet terminus generalis, quoniam ex ea, loco $n$ numeros omnes successive substituendo, omnes Seriei termini nascuntur. Deinde ex fractione $\frac{A}{(1-pz)^2}$ oritur Series $A+2Apz + 3Ap^2z^2 + 4Ap^3z^3 + \&c.$, cujus terminus generalis est $(n+1)Ap^n z^n$. Tum ex fractione $\frac{A}{(1-pz)^3}$ oritur Series $A + 3Apz + 6Ap^2z^2 + 10Ap^3z^3 + \&c.$, cujus terminus generalis est $\frac{(n+1)(n+2)}{1 \cdot 2} Ap^n z^n$. Generatim autem fractio $\frac{A}{(1-pz)^k}$ prabet Seriem hanc $A+kApz + \frac{k(k+1)}{1 \cdot 2} Ap^2z^2 + \frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3} Ap^3z^3 + \&c.$, cujus terminus generalis est $\frac{(n+1)(n+2)(n+3) \ldots (n+k-1)}{1 \cdot 2 \cdot 3 \ldots (k-1)} Ap^n z^n$. Ex ipsa autem Seriei progressione colligitur hic idem terminus = $\frac{k(k+1)(k+2) \ldots (k+n-1)}{1 \cdot 2 \cdot 3 \ldots n} Ap^n z^n$: hæc vero expressio Euleri Introduct. in Anal. infin. parv. Z