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フィボナッチ数列の逆数和の連分数展開

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ランベルト級数の連分数展開の結果を用いて、フィボナッチ数列の逆数和 $ψ$ を連分数展開する。

$\begin{array}{r} \sum_{n = 1}^{\infty} \frac{z^{n}}{r^{n} - a q^{n}} = K_{n = 1}^{\infty} \frac{u_{n}}{r^{n} - a q^{n}} \end{array}$
${\begin{aligned} u_{1} & = z \\ u_{2 n} & = - (q r)^{n - 1} {(r^{n} - a q^{n})}^{2} z \\ u_{2 n + 1} & = - a (q r)^{n} {(r^{n} - q^{n})}^{2} z \end{aligned}$

$\begin{array}{rcl} (左辺) & = & \frac{z}{r} \sum_{n = 0}^{\infty} \frac{{(\frac{z}{r})}^{n}}{1 - \frac{a q}{r} {(\frac{q}{r})}^{n}} \\ = & \frac{\frac{z}{r}}{1 - \frac{a q}{r} - \frac{(1 - \frac{a q}{r})^{2} \frac{z}{r}}{1 - \frac{a q^{2}}{r^{2}} - \frac{\frac{a q}{r} (1 - \frac{q}{r})^{2} \frac{z}{r}}{1 - \frac{a q^{3}}{r^{3}} - ⋱}}} \\ = & \frac{z}{r - a q - \frac{(r - a q)^{2} z}{r^{2} - a q^{2} - \frac{a q r (r - q)^{2} z}{r^{3} - a q^{3} - ⋱}}} \\ = & (右辺) \end{array}$

$F_{n}$ :n番目のフィボナッチ数
$\begin{aligned} \sum_{n = 1}^{\infty} \frac{z^{n}}{F_{n}} & = \frac{z}{F_{1} + K_{n = 2}^{\infty} \frac{(- 1)^{⌊ \frac{n + 1}{2} ⌋} F_{⌊ \frac{n}{2} ⌋}^{2} z}{F_{n}}} \\ = \frac{z}{1 - \frac{1^{2} z}{1 + \frac{1^{2} z}{2 + \frac{1^{2} z}{3 - \frac{1^{2} z}{5 - \frac{2^{2} z}{8 + \frac{2^{2} z}{13 + ⋱}}}}}}} \end{aligned}$

$\begin{array}{rcl} ϕ_{\pm} & := & \frac{1 \pm \sqrt{5}}{2} \\ ϕ_{+} ϕ_{-} & = & - 1 \\ F_{n} & = & \frac{ϕ_{+}^{n} - ϕ_{-}^{n}}{\sqrt{5}} \\ (左辺) & = & \sqrt{5} \sum_{n = 1}^{\infty} \frac{z^{n}}{ϕ_{+}^{n} - ϕ_{-}^{n}} \\ = & K_{n = 1}^{\infty} \frac{u_{n}}{ϕ_{+}^{n} - ϕ_{-}^{n}} ∵ 定理1 \\ = & K_{n = 1}^{\infty} \frac{u_{n}}{\sqrt{5} F_{n}} \\ = & (右辺) \end{array}$
${\begin{aligned} u_{1} & = \sqrt{5} z \\ u_{2 n} & = - (ϕ_{+} ϕ_{-})^{n - 1} {(ϕ_{+}^{n} - ϕ_{-}^{n})}^{2} z & = 5 (- 1)^{n} F_{n}^{2} z \\ u_{2 n + 1} & = - (ϕ_{+} ϕ_{-})^{n} {(ϕ_{+}^{n} - ϕ_{-}^{n})}^{2} z & = 5 (- 1)^{n + 1} F_{n}^{2} z \end{aligned}$