Introductio in analysin infinitorum

EVOLUTIONE FACTORUM ORTIS. 227 CAP. XV. $Q=(1+\frac{1}{2^n})(1+\frac{1}{3^n})(1+\frac{1}{5^n})(1+\frac{1}{7^n})(1+\frac{1}{11^n}) \text{&c.}$ erit $P=1-\frac{1}{2^n}-\frac{1}{3^n}+\frac{1}{4^n}-\frac{1}{5^n}-\frac{1}{6^n}-\frac{1}{7^n}+\frac{1}{8^n}-\frac{1}{9^n}+\text{&c.}$ $Q=1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{5^n}+\frac{1}{6^n}+\frac{1}{7^n}+\frac{1}{10^n}+\frac{1}{11^n}+\text{&c.}$ similique modo habebitur $PQ=1$. Cognita ergo alterius Seriei summa, simul alterius innotescet. 277. Vicissim porro ex cognitis summis harum Serierum, assignari poterunt valores Factorum infinitorum. Sit nimirum $M=1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5 [TRUNCATED]