188.

It now remains to prove that the sum $Φ = \sum_{1}^{\infty} (w_{n} - ζ w_{n - 1}) \log w_{n} - [(1 - ζ) n - 1] w_{n} \log ζ, \Label [e q n] (356) \upshape (356)$ 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 $ζ$ , makes the sum $Φ$ a minimum. In this case $δ Φ = 0$ , or

$\sum_{1}^{\infty} (δ w_{n} - ζ δ w_{n - 1}) \log w_{n} + (w_{n} - ζ w_{n - 1}) \frac{δ w_{n}}{w_{n}} \Label [e q n] (357) \upshape (357) - [(1 - ζ) n - 1] δ w_{n} \log ζ = 0,$

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

The solution of this functional equation is

$w_{n} = (1 - ζ) ζ^{n - 1} \Label [e q n] (361) \upshape (361)$

for it satisfies [eqn:(360)] (360) as well as [eqn:(353)] (353) and [eqn:(354)] (354). With this value [eqn:(356)] (356) becomes

$Φ = 0 . \Label [e q n] (362) \upshape (362)$

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