Introductio in analysin infinitorum

EVOLUTIONE FACTORUM ORTIS. 233 in his Factoribus numeratores unitate superant denominatores, simul vero sumti praebent bi-quadrata numerorum primorum imparium $3, 5, 7, 11,$ &c. CAP. XV. 278. Quoniam hic summam Seriei $M=1+\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}+\frac{1}{5^n}+\frac{1}{6^n}+\,&c.$ ad Factores reduximus, ad Logarithmos commode progredi licebit. Nam, cum fit $M=\frac{1}{(1-\frac{1}{2^n})(1-\frac{1}{3^n})(1-\frac{1}{5^n})(1-\frac{1}{7^n})(1-\frac{1}{11^n})\,&c.}$, erit $lM=-l(1-\frac{1}{2^n})-l(1-\frac{3}{3^n})-l(1-\frac{1}{5^n})-l(1-\frac{1}{7^n})-\,&c.$ Hinc, sumendis Logarithmis hyperbolicis, erit $lM = +1(\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{5^n}+\frac{1}{7^n}+\frac{1}{11^n}+\,&c.)$ $+\frac{1}{2}(\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\frac{1}{5^{2n}}+\frac{1}{7^{2n}}+\frac{1}{11^{2n}}+\,&c.)$ $+\frac{1}{3}(\frac{1}{2^{3n}}+\frac{1}{3^{3n}}+\frac{1}{5^{3n}}+\frac{1}{7^{3n}}+\frac{1}{11^{3n}}+\,&c.)$ $+\frac{1}{4}(\frac{1}{2^{4n}}+\frac{1}{3^{4n}}+\frac{1}{5^{4n}}+\frac{1}{7^{4n}}+\frac{1}{11^{4n}}+\,&c.)$ $&c.$ Quod si insuper ponamus Euleri *Introduct. in Anal. infin. parv.* Gg N=