Page:Annales de mathématiques pures et appliquées, 1814-1815, Tome 5.djvu/244

238

ANALISE

${\begin{array}{lrll}\Lambda (1+n)&=\Lambda 1&-\operatorname {Log} .\left[1-{\frac {n}{(1+p)(1+n)}}\right]&-\Lambda {\frac {1}{1+p}}\\&+1&+n+\ldots &-1\\&+{\frac {1}{2}}&-{\frac {n}{4}}+{\frac {n^{2}}{8}}-{\frac {n^{3}}{16}}+\ldots &-{\frac {1}{2}}\\&&+{\frac {n}{2}}-{\frac {n^{2}}{4}}+{\frac {n^{3}}{8}}-\ldots &\\&+{\frac {1}{3}}&-{\frac {2n}{9}}+{\frac {4n^{2}}{27}}-{\frac {8n^{3}}{81}}+\ldots &-{\frac {1}{3}}\\&&+{\frac {n}{3}}-{\frac {2n^{2}}{9}}+{\frac {4n^{3}}{27}}-\ldots &\\\end{array}}$

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

${\begin{array}{rrl}+{\frac {1}{p}}&-{\frac {(p-1)n}{p^{2}}}&+{\frac {(p-1)^{2}n^{2}}{p^{2}}}-{\frac {(p-1)^{3}n^{3}}{p^{4}}}+\ldots -{\frac {1}{p}}\\&+{\frac {n}{p}}&-{\frac {(p-1)n^{2}}{p^{2}}}+{\frac {(p-1)^{2}n^{3}}{p^{3}}}-\ldots \\&-n&+{\frac {n^{2}}{2}}-{\frac {n^{3}}{3}}+\ldots \\\end{array}}$

$p$ étant arbitraire, faisons-le infini ; $\operatorname {Log} .\left[1-{\frac {n}{(1+p)(1+n)}}\right]$ et $\Lambda {\frac {1}{1+p}}$ disparaîtront, et il restera

$\Lambda (1+n)=\Lambda 1+n\left({\tfrac {1}{4}}+{\tfrac {1}{9}}+{\tfrac {1}{16}}+\ldots \right)+n^{2}\left(+{\tfrac {1}{2}}-{\tfrac {1}{8}}-{\tfrac {1}{27}}-{\tfrac {1}{64}}-\ldots \right)$

$+n^{3}\left(-{\tfrac {1}{3}}+{\tfrac {1}{16}}+{\tfrac {4}{81}}+\ldots \right)+n^{4}\left(+{\tfrac {1}{4}}-{\tfrac {1}{32}}-\ldots \right)+\ldots \,;$

équation que nous écrirons ainsi :

$\Lambda (1+n)=\Lambda 1+\lambda _{1}n+\lambda _{2}n^{2}+\lambda _{3}n^{3}+\ldots \qquad$(E)

3. Or, si l’on fait

$S_{1}=1+{\frac {1}{2^{\varepsilon }}}+{\frac {1}{3^{\varepsilon }}}+{\frac {1}{4^{\varepsilon }}}+\ldots ,$

$\varepsilon$ étant employé comme indice général, on trouvera