188.

It now remains to prove that the sum $\Phi ={\sum }_1^{\infty }(w_n-\zeta w_{n-1})\log w_n-[(1-\zeta )n-1]w_n\log \zeta ,\mathrm{\backslash Label}[eqn](356)\text{upshape (356)}$ where the quantities $w_n$ are subject only to the restrictions that [eqn:(353)] (353) and [eqn:(354)] (354) can never become negative. For this purpose we determine that system of values of the $w$'s which, with a fixed value of $\zeta$, makes the sum $\Phi$ a minimum. In this case $\delta \Phi =0$, or

$$\begin{gathered} {\sum }_1^{\infty }(\delta w_n-\zeta \delta w_{n-1})\log w_n+(w_n-\zeta w_{n-1})\frac{\delta w_n}{w_n}\mathrm{\backslash Label}[eqn](357)\text{upshape (357)} \\ -[(1-\zeta )n-1]\delta w_n\log \zeta =0, \end{gathered}$$

where, according to [eqn:(353)] (353) and [eqn:(354)] (354), ${\sum }_1^{\infty }\delta w_n=0\text{and}\delta w_0=0.\mathrm{\backslash Label}[eqn](358)\text{upshape (358)}$ If we suppose all the separate terms of the sum to be written out, the equation may be put into the following form: ${\sum }_1^{\infty }\delta w_n\{\log w_n-\zeta \log w_{n+1}+\frac{w_n-\zeta w_{n-1}}{w_n}-[(1-\zeta )n-1]\log \zeta \}=0.\mathrm{\backslash Label}[eqn](359)\text{upshape (359)}$ From this, by taking account of [eqn:(358)] (358), we get as the condition for a minimum, that $\log w_n-\zeta \log w_{n+1}+\frac{w_n-\zeta w_{n-1}}{w_n}-[(1-\zeta )n-1]\log \zeta \mathrm{\backslash Label}[eqn](360)\text{upshape (360)}$ must be independent of $n$.

The solution of this functional equation is $w_n=(1-\zeta ){\zeta }^{n-1}\mathrm{\backslash Label}[eqn](361)\text{upshape (361)}$ for it satisfies [eqn:(360)] (360) as well as [eqn:(353)] (353) and [eqn:(354)] (354). With this value [eqn:(356)] (356) becomes $\Phi =0.\mathrm{\backslash Label}[eqn](362)\text{upshape (362)}$