Introductio in analysin infinitorum

LIB. I. 188 DE SERIEBUS si fuerit terminus generalis futurus fit $n = 6m+0$ $(\frac{nn}{12} + \frac{n}{2} + 1) z^n$ $n = 6m+1$ $(\frac{nn}{12} + \frac{n}{2} + \frac{5}{12}) z^n$ $n = 6m+2$ $(\frac{nn}{12} + \frac{n}{2} + \frac{2}{3}) z^n$ $n = 6m+3$ $(\frac{nn}{12} + \frac{n}{2} + \frac{3}{4}) z^n$ $n = 6m+4$ $(\frac{nn}{12} + \frac{n}{2} + \frac{2}{3}) z^n$ $n = 6m+5$ $(\frac{nn}{12} + \frac{n}{2} + \frac{5}{12}) z^n$ Sic, si fuerit $n=50$, valet forma $n=6m+2$, eritque terminus Seriei $= 2342\frac{5}{6}$. EXEMPLUM II. Ex fractione $\frac{1+z+zz}{1-z-z^4+z^5}$ oritur [TRUNCATED]