Introductio in analysin infinitorum

232 DE SERIEBUS EX LIB. I. EXEMPLUM III. Quia ex superioribus valores ipsius M tantum sin sit numerus par, assignare licet, ponamus $n=4$, eritque $M=1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\&c. = \frac{\pi^4}{90}$ $N=1+\frac{1}{2^8}+\frac{1}{3^8}+\frac{1}{4^8}+\frac{1}{5^8}+\frac{1}{6^8}+\&c. = \frac{\pi^8}{9450}$ Hinc primae sequentes Series summantur $\frac{90}{\pi^4} = 1 - \frac{1}{2^4} - \frac{1}{3^4} - \frac{1}{5^4} + \frac{1}{6^4} - \frac{1}{7^4} + \frac{1}{10^4} - \frac{1}{11^4} \&c.$ $\frac{9450}{\pi^8} = 1 - \frac{1}{2^8} - \frac{1}{3^8} - \frac{1}{5^8} + \frac{1}{6^8} - \frac{1}{7^8 [TRUNCATED]