Introductio in analysin infinitorum

EVOLUTIONE FACTORUM ORTIS. 235 CAP. XV. 279. Si $n=1$ erit $M=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\&c. = l\infty$, & $N=\frac{\pi\pi}{6}$; hincque erit $l.l\infty - \frac{1}{2}l\frac{\pi\pi}{6} =$ $+ 1(\frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \&c.)$ $+ \frac{1}{3}(\frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \frac{1}{11^3} + \&c.)$ $+ \frac{1}{5}(\frac{1}{2^5} + \frac{1}{3^5} + \frac{1}{5^5} + \frac{1}{7^5} + \frac{1}{11^5} + \&c.)$ $+ \frac{1}{7}(\frac{1}{2^7} + \frac{1}{3^7} + \frac{1}{5^7} + \frac{1}{7 [TRUNCATED]