Pagina:Werkecarlf03gausrich.djvu/159

Generaliter autem pro quovis valore ipsius ${z}$ habemus

[41]	$\Pi (k,z+1)=\Pi (k,z)\cdot {\frac {1+z}{1+{\frac {1+z}{k}}}}$
[42]	$\Pi (k+1,z)=\Pi (k,z)\cdot \left\{{\frac {\left(1+{\frac {1}{k}}\right)^{z+1}}{1+{\frac {1+z}{k}}}}\right\}$

adeoque, quum $\Pi (1,z)={\frac {1}{z+1}},$

[43]	$\Pi (k,z)={\frac {1}{z+1}}\cdot {\frac {2^{z+1}}{1^{z}\cdot (2+z)}}\cdot {\frac {3^{z+1}}{2^{z}(3+z)}}\cdot {\frac {4^{z+1}}{3^{z}(4+z)}}\cdots {\frac {k^{z+1}}{(k-1)^{z}(k+z)}}$

20.

Imprimis vero attentione dignus est limes, ad quem pro valore dato ipsius $z$ functio $\Pi (k,z)$ continuo converget, dum $k$ in infinitum crescit. Sit primo $h$ valor finitus ipsius $k$ maior quam $z,$ patetque, si $k$ transire supponatur ex $h$ in $h+1,$ logarithmum ipsius $\Pi (k,z)$ accipere incrementum, quod per seriem convergentem sequentem exprimatur

${\frac {z(1+z)}{2(h+1)^{2}}}+{\frac {z(1-zz)}{3(h+1)^{3}}}+{\frac {z(1+z^{3})}{4(h+1)^{4}}}+{\frac {z(1-z^{4})}{5(h+1)^{5}}}+{\text{etc.}}$

Si itaque $k$ e valore $h$ transit in $h+n,$ logarithmus ipsius $\Pi (k,z)$ accipiet incrementum

${\begin{alignedat}{3}&&&{\tfrac {1}{2}}z(1+z)&\left({\frac {1}{(h+1)^{2}}}+{\frac {1}{(h+2)^{2}}}+{\frac {1}{(h+3)^{2}}}+{\text{etc.}}+{\frac {1}{(h+n)^{2}}}\right)\\&+&{\tfrac {1}{3}}&z(1-zz)&\left({\frac {1}{(h+1)^{3}}}+{\frac {1}{(h+2)^{3}}}+{\frac {1}{(h+3)^{3}}}+{\text{etc.}}+{\frac {1}{(h+n)^{3}}}\right)\\&+&{\tfrac {1}{4}}&z(1+z^{3})&\left({\frac {1}{(h+1)^{4}}}+{\frac {1}{(h+2)^{4}}}+{\frac {1}{(h+3)^{4}}}+{\text{etc.}}+{\frac {1}{(h+n)^{4}}}\right)\\&+&&{\text{etc.}}&\end{alignedat}}$

quod semper finitum manere, etiamsi $n$ in infinitum crescat, facile demonstrari potest. Quare nisi iam in $\Pi (h,z)$ factor infinitus affuerit, i. e nisi $z$ sit numerus integer negativus, limes ipsius $\Pi (k,z)$ pro $k=\infty$ certo erit quantitas finita. Manifesto itaque $\Pi (\infty ,z)$ tantummodo a $z$ pendet, sive functionem ipsius $z$ ex asse determinatam exhibet, quae abhinc simpliciter per $\Pi z$ denotabitur. Definimus itaque functionem $\Pi z$ per valorem producti

${\frac {1\;.\;2\;.\;3\;.\ldots .\;k\;.\;k^{z}}{(z+1)(z+2)(z+3)\ldots (z+k)}}$