Introductio in analysin infinitorum

240 DE SERIEBUS EX LIB. I. cum hic tantum numeri impares occurrunt, eos qui sint formæ $4m+1$ habere signum $+$, reliquos formæ $4m-1$ signum $-$. Sit igitur $A=1-\frac{1}{3^n}+\frac{1}{5^n}-\frac{1}{7^n}+\frac{1}{9^n}-\frac{1}{11^n}+\frac{1}{13^n}-\frac{1}{15^n}+$ &c. $\frac{1}{3^n}A=\frac{1}{3^n}-\frac{1}{9^n}+\frac{1}{15^n}-\frac{1}{21^n}+\frac{1}{27^n}-$ &c., addatur, erit $(1+\frac{1}{3^n})A=1+\frac{1}{5^n}-\frac{1}{7^n}-\frac{1}{11^n}+\frac{1}{13^n}+\frac{1}{17^n}+$ &c., $=B$ $\frac{1}{5^n}B=\frac{1}{5^n}+\frac{1}{25^n}-\frac{1}{35^n}-\frac{1}{55^n}-$ &c., subtrahatur, erit $(1-\frac{1}{5^n [TRUNCATED]