零数列による級数の加速

[0]和、スカラー倍は閉じている。
\begin{eqnarray} \forall \{w_{n}\}_{n\in\mathbb{N}_{0}},\{\overline{w}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(w_{n}+\overline{w}_{n})a_{n}&=&\sum_{n=0}^{\infty}w_{n}a_{n}+\sum_{n=0}^{\infty}\overline{w}_{n}a_{n}\\ &=&0+0\\ &=&0 \end{eqnarray}
また
\begin{eqnarray} \forall c\in\mathbb{C}:\forall \{w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(cw_{n})a_{n}&=&c\sum_{n=0}^{\infty}w_{n}a_{n}\\ &=&c0\\ &=&0 \end{eqnarray}
[1]結合法則
\begin{eqnarray} \forall \{w_{n}\}_{n\in\mathbb{N}_{0}},\{\overline{w}\}_{n\in\mathbb{N}_{0}},\{\overline{\overline{w}}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(w_{n}+(\overline{w}_{n}+\overline{\overline{w}}_{n}))a_{n}&=&\sum_{n=0}^{\infty}((w_{n}+\overline{w}_{n})+\overline{\overline{w}}_{n})a_{n} \end{eqnarray}
[2]単位元：$\{0\}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}})$
\begin{eqnarray} \forall \{w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(w_{n}+0)a_{n}&=&\sum_{n=0}^{\infty}(0+w_{n})a_{n}\\ &=&\sum_{n=0}^{\infty}w_{n}a_{n} \end{eqnarray}
[3]逆元：$\{-w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}})$
\begin{eqnarray} \forall \{w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(w_{n}+(-w_{n}))a_{n}&=&\sum_{n=0}^{\infty}(-w_{n}+w_{n})a_{n}\\ &=&\sum_{n=0}^{\infty}0a_{n}\\ &=&0 \end{eqnarray}
[4]可換性
\begin{eqnarray} \forall \{w_{n}\}_{n\in\mathbb{N}_{0}},\{\overline{w}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(w_{n}+\overline{w}_{n})a_{n}&=&\sum_{n=0}^{\infty}(\overline{w}_{n}+w_{n})a_{n} \end{eqnarray}
[5]ベクトルに関する分配法則
\begin{eqnarray} \forall c\in\mathbb{C}:\forall \{w_{n}\}_{n\in\mathbb{N}_{0}},\{\overline{w}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(c(w_{n}+\overline{w}_{n}))a_{n}=\sum_{n=0}^{\infty}(cw_{n}+c\overline{w}_{n})a_{n} \end{eqnarray}
[6]スカラーに関する分配法則
\begin{eqnarray} \forall c,d\in\mathbb{C}:\forall \{w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}((c+d)w_{n})a_{n}=\sum_{n=0}^{\infty}(cw_{n}+dw_{n})a_{n} \end{eqnarray}
[7]スカラー積に関する結合法則
\begin{equation} \forall c,d\in\mathbb{C}:\forall \{w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}((cd)w_{n})a_{n}=\sum_{n=0}^{\infty}(c(dw_{n}))a_{n} \end{equation}
[8]スカラー積の単位元：$1\in\mathbb{C}$
\begin{eqnarray} \forall \{w_{n}\}_{n\in\mathbb{N}_{0}}\in Zero(\{a_{n}\}_{n\in\mathbb{N}_{0}}):\sum_{n=0}^{\infty}(1w_{n})a_{n}=\sum_{n=0}^{\infty}w_{n}a_{n} \end{eqnarray}

そのまま計算するだけ。
\begin{eqnarray} \sum_{n=0}^{\infty}(1+w_{n})a_{n}&=&\sum_{n=0}^{\infty}a_{n}+\sum_{n=0}^{\infty}w_{n}a_{n}\\ &=&S(\{a_{n}\}_{n\in\mathbb{N}_{0}})+0\\ &=&S(\{a_{n}\}_{n\in\mathbb{N}_{0}}) \end{eqnarray}

[0]$I_{n}\coloneqq \int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx$とおく。
[1]すると$I_{n}$は$n$に関して単調減少。なぜなら、$\forall x\in(0,1):\sin{x}\in(0,1)$なので$\forall n\in\mathbb{N}_{0}:\forall x\in(0,1):0\lt \sin^{n+1}{x}\lt\sin^{n}{x}$。
ゆえに、$0\lt I_{n+1}\lt I_{n}$
[2]下記の様に計算ができる
\begin{eqnarray} I_{n}&=&\int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx\\ &=&-\sin^{n-1}{x}\cos{x}|_{0}^{\frac{\pi}{2}}+(n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}\cos^{2}{x}dx\\ &=&(n-1)I_{n-2}-(n-1)I_{n} \end{eqnarray}
\begin{equation} I_{n}=\frac{n-1}{n}I_{n-2} \end{equation}
[3][2]を用いると以下の結果を得る。
\begin{eqnarray} \left\{ \begin{array}{l} I_{2n}=\frac{2n-1}{2n}\frac{2n-3}{2n-2}\cdots\frac{1}{2}\frac{\pi}{2}=\frac{(2n)!}{2^{2n}(n!)^{2}}\frac{\pi}{2}\\ I_{2n+1}=\frac{2n}{2n+1}\frac{2n-2}{2n-1}\cdots\frac{2}{3}=\frac{2^{2n}(n!)^{2}}{(2n+1)!} \end{array} \right. \end{eqnarray}
[4][1]より$I_{2n+1}\lt I_{2n}\lt I_{2n-1}$とする。すると下記の様に計算できる。
\begin{eqnarray} 1\lt\frac{I_{2n}}{I_{2n-1}}=n\{\frac{(2n)!}{2^{2n}(n!)^{2}}\}^{2}\pi\lt\frac{1}{1+\frac{1}{2n}} \end{eqnarray}
[5]
\begin{equation} \pi=\lim_{n\rightarrow\infty}\frac{2^{2n}(n!)^{4}}{n\{(2n)!\}^{2}} \end{equation}

[1]Euler-Maclaurin展開より下記の様に計算できる。
\begin{eqnarray} \log{N!}&=&\sum_{n=0}^{N-1}\log{(n+1)}\\ &=&\int_{0}^{N-1}\log{(x+1)}dx+\frac{\log{N}}{2}+\frac{\frac{1}{N}-1}{12}+\cdots\\ &=&\log{C}+N\log{N}-N+\frac{\log{N}}{2}+O(\frac{1}{N})\\ &=&\log{(CN^{N+\frac{1}{2}}e^{-N})}+O(\frac{1}{N}) \end{eqnarray}
[2]上記の計算により
\begin{equation} C=\lim_{n\rightarrow\infty}\frac{N!e^{N}}{N^{N+\frac{1}{2}}} \end{equation}
[3]定数$C$についてはWallisの公式より下記の様に計算できる。
\begin{eqnarray} \sqrt{\pi}&=&\lim_{n\rightarrow\infty}\frac{2^{2n}(n!)^{2}}{n^{\frac{1}{2}}\{(2n)!\}}\\ &=&\frac{1}{\sqrt{2}}\lim_{n\rightarrow\infty}\frac{n!e^{n}}{n^{n+\frac{1}{2}}}\frac{n!e^{n}}{n^{n+\frac{1}{2}}}\frac{(2n)^{2n+\frac{1}{2}}}{(2n)!}\\ &=&\frac{C}{\sqrt{2}} \end{eqnarray}
[4]ゆえに$C=\sqrt{2\pi}$が得られるので、最終的に以下の式が証明された。
\begin{equation} N!=\frac{\sqrt{2\pi N}N^{N}}{e^{N}}+e^{O(\frac{1}{N})} \end{equation}

\begin{eqnarray} \begin{pmatrix}2n\\n\end{pmatrix}&=&\frac{(2n)!}{n!n!}\\ &=&\frac{\sqrt{4\pi n}(2n)^{2n}}{e^{2n}}\frac{e^{n}}{\sqrt{2\pi n}n^{n}}\frac{e^{n}}{\sqrt{2\pi n}n^{n}}\\ &=&\frac{4^{n}}{\sqrt{\pi n}} \end{eqnarray}

[1]$t_{n}\coloneqq\frac{1}{n^{p}\begin{pmatrix}2n\\n\end{pmatrix}^{q}}$とおく。
[2]$t_{n}$は超幾何項。
\begin{eqnarray} \frac{t_{n+1}}{t_{n}}&=&\frac{n^{p}\begin{pmatrix}2n\\n\end{pmatrix}^{q}}{(n+1)^{p}\begin{pmatrix}2n+2\\n+1\end{pmatrix}^{q}}\\ &=&\frac{n^{p}(n+1)^{2q}}{(n+1)^{p}(2n+2)^{q}(2n+1)^{q}}\\ &=&\frac{n^{p}(n+1)^{q-p}}{2^{q}(2n+1)^{q}}\\ &=&\frac{n^{p}(n+1)^{q-p}}{2^{2q+1}(n+\frac{1}{2})^{q}} \end{eqnarray}
[3]ゆえに
\begin{eqnarray} t_{n}&=&\frac{(n-1)^{p}n^{q-p}}{2^{2q+1}(n-\frac{1}{2})^{q}}t_{n-1}\\ &=&\frac{(1)^{p}_{n-1}(2)^{q-p}_{n-1}}{2^{2q(n-1)}(\frac{3}{2})^{q}_{n-1}}t_{1}\\ &=&\frac{(1)^{p}_{n-1}(2)^{q-p}_{n-1}}{2^{q(2n-1)}(\frac{3}{2})^{q}_{n-1}} \end{eqnarray}
[4]以上の事を代入すると
\begin{eqnarray} S(p,q)&=&\sum_{n=1}^{\infty}t_{n}\\ &=&\sum_{n=0}^{\infty}\frac{(1)^{p}_{n}(2)^{q-p}_{n}}{2^{q(2n+1)}(\frac{3}{2})^{q}_{n}}\\ &=&\frac{1}{2^{q}}\sum_{n=0}^{\infty}\frac{(1)^{p+1}_{n}(2)^{q-p}_{n}}{(\frac{3}{2})^{q}_{n}}\frac{1}{n!4^{qn}}\\ &=&\frac{1}{2^{q}}{}_{q+1}F_{q}(\{1\}^{p+1},\{2\}^{q-p};\{\frac{3}{2}\}^{q};\frac{1}{4^{q}}) \end{eqnarray}

実際に計算すればよい。
\begin{eqnarray} (a)_{2n}&=&a(a+1)(a+2)\cdots(a+2n-2)(a+2n-1)\\ &=&a(a+2)(a+4)\cdots(a+2n-2)(a+1)(a+3)\cdots(a+2n-1)\\ &=&2^{2n}\frac{a}{2}(\frac{a}{2}+1)(\frac{a}{2}+2)\cdots(\frac{a}{2}+n-1)(\frac{a}{2}+\frac{1}{2})(\frac{a}{2}+\frac{1}{2}+1)\cdots(\frac{a}{2}+\frac{1}{2}+n-1)\\ &=&2^{2n}(\frac{a}{2})_{n}(\frac{a+1}{2})_{n} \end{eqnarray}
\begin{eqnarray} (a)_{2n+1}=2^{2n}a(\frac{a+1}{2})_{n}(\frac{a}{2}+1)_{n} \end{eqnarray}

これも実際に計算すればいい。
\begin{eqnarray} (a)_{3n}&=&a(a+3)(a+6)\cdots(a+3n-3)(a+1)(a+4)\cdots(a+3n-2)(a+2)(a+5)\cdots(a+3n-1)\\ &=&3^{3n}\frac{a}{3}(\frac{a}{3}+1)(\frac{a}{3}+2)\cdots(\frac{a}{3}+n-1)\frac{a+1}{3}(\frac{a+1}{3}+1)\cdots(\frac{a+1}{n}+n-1)\frac{a+2}{3}(\frac{a+2}{3}+1)\cdots(\frac{a+2}{3}+n-1)\\ &=&3^{3n}(\frac{a}{3})_{n}(\frac{a+1}{3})_{n}(\frac{a+2}{3})_{n} \end{eqnarray}
\begin{eqnarray} (a)_{3n+1}=3^{n}a(\frac{a+1}{3})_{n}(\frac{a+2}{3})_{n}(\frac{a}{3}+1)_{n} \end{eqnarray}
\begin{eqnarray} (a)_{3n+2}=3^{n}a(a+1)(\frac{a+2}{3})_{n}(\frac{a}{3}+1)_{n}(\frac{a+1}{3}+1)_{n} \end{eqnarray}

[1]
\begin{eqnarray} {}_{p}F_{q+1}(a_{1},a_{2},...,a_{p};b_{1},b_{2},...,b_{q};z)+{}_{p}F_{q}(a_{1},a_{2},...,a_{p};b_{1},b_{2},...,b_{q};-z)&=&2\sum_{n=0}^{\infty}\frac{(a_{1})_{2n}(a_{2})_{2n}\cdots(a_{p})_{2n}}{(b_{1})_{2n}(b_{2})_{2n}\cdots(b_{q})_{2n}}\frac{z^{2n}}{(2n)!}\\ &=&2\sum_{n=0}^{\infty}2^{2(p-q)}\frac{(\frac{a_{1}}{2})_{n}(\frac{a_{1}+1}{2})_{n}(\frac{a_{2}}{2})_{n}(\frac{a_{2}+1}{2})_{n}\cdots(\frac{a_{p}}{2})_{n}(\frac{a_{p}+1}{2})_{n}}{(\frac{b_{1}}{2})_{n}(\frac{b_{1}+1}{2})_{n}(\frac{b_{2}}{2})_{n}(\frac{a_{b}+1}{2})_{n}\cdots(\frac{b_{q}}{2})_{n}(\frac{b_{q}+1}{2})_{n}}\frac{z^{2n}}{2^{2n}(\frac{1}{2})_{n}n!}\\ &=&2{}_{2p}F_{2q+1}(\frac{a_{1}}{2},\frac{a_{1}+1}{2},\frac{a_{2}}{2},\frac{a_{2}+1}{2},...,\frac{a_{p}}{2},\frac{a_{p}+1}{2};\frac{b_{1}}{2},\frac{b_{1}+1}{2},\frac{b_{2}}{2},\frac{b_{2}+1}{2},...,\frac{b_{q}}{2},\frac{b_{q}+1}{2},\frac{1}{2};(2^{p-q-1}z)^{2}) \end{eqnarray}
[2]
\begin{eqnarray} {}_{p}F_{q}(a_{1},a_{2},...,a_{p};b_{1},b_{2},...,b_{q};z)-{}_{p}F_{q}(a_{1},a_{2},...,a_{p};b_{1},b_{2},...,b_{q};-z)&=&2\sum_{n=0}^{\infty}\frac{(a_{1})_{2n+1}(a_{2})_{2n+1}\cdots(a_{p})_{2n+1}}{(b_{1})_{2n+1}(b_{2})_{2n+1}\cdots(b_{q})_{2n+1}}\frac{z^{2n+1}}{(2n+1)!}\\ &=&2z\sum_{n=0}^{\infty}\frac{(a_{1})_{2n+1}(a_{2})_{2n+1}\cdots(a_{p})_{2n+1}}{(b_{1})_{2n+1}(b_{2})_{2n+1}\cdots(b_{q})_{2n+1}}\frac{z^{2n}}{2^{2n}(\frac{3}{2})_{n}n!}\\ &=&2z\sum_{n=0}^{\infty}2^{2(p-q)n}\frac{a_{1}a_{2}\cdots a_{p}(\frac{a_{1}+1}{2})_{n}(\frac{a_{1}}{2}+1)_{n}(\frac{a_{2}+1}{2})_{n}(\frac{a_{2}}{2}+1)_{n}\cdots(\frac{a_{p}+1}{2})_{n}(\frac{a_{p}}{2}+1)_{n}}{b_{1}b_{2}\cdots b_{q}(\frac{b_{1}+1}{2})_{n}(\frac{b_{1}}{2}+1)_{n}(\frac{b_{2}+1}{2})_{n}(\frac{b_{2}}{2}+1)_{n}\cdots(\frac{b_{q}+1}{2})_{n}(\frac{b_{q}}{2}+1)_{n}(\frac{3}{2})_{n}}\frac{1}{n!}(\frac{z}{2})^{2n}\\ &=&2\frac{a_{1}a_{2}\cdots a_{p}}{b_{1}b_{2}\cdots b_{q}}z\sum_{n=0}^{\infty}\frac{(\frac{a_{1}+1}{2})_{n}(\frac{a_{1}}{2}+1)_{n}(\frac{a_{2}+1}{2})_{n}(\frac{a_{2}}{2}+1)_{n}\cdots(\frac{a_{p}+1}{2})_{n}(\frac{a_{p}}{2}+1)_{n}}{(\frac{b_{1}+1}{2})_{n}(\frac{b_{1}}{2}+1)_{n}(\frac{b_{2}+1}{2})_{n}(\frac{b_{2}}{2}+1)_{n}\cdots(\frac{b_{q}+1}{2})_{n}(\frac{b_{q}}{2}+1)_{n}(\frac{3}{2})_{n}}\frac{1}{n!}(2^{p-q-1}z)^{2n}\\ &=&2\frac{a_{1}a_{2}\cdots a_{p}}{b_{1}b_{2}\cdots b_{q}}z{}_{2p}F_{2q+1}(\frac{a_{1}+1}{2},\frac{a_{1}+2}{2},\frac{a_{2}+1}{2},\frac{a_{2}+2}{2},...,\frac{a_{p}+1}{2},\frac{a_{p}+2}{2};\frac{b_{1}+1}{2},\frac{b_{1}+2}{2},\frac{b_{2}+1}{2},\frac{b_{2}+2}{2},...,\frac{b_{q}+1}{2},\frac{b_{q}+2}{2},\frac{3}{2};(2^{p-q-1}z)^{2}) \end{eqnarray}

[1]$f(x)=\frac{1}{\sqrt{1-x}}$とおくと$f^{(n)}(x)=\frac{(\frac{1}{2})_{n}}{(1-x)^{\frac{2n+1}{2}}}$なので
\begin{equation} f(x)=\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_{n}}{n!}x^{n} \end{equation}
[2]ゆえにこれを用いると
\begin{eqnarray} K(k)&=&\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_{n}}{n!}k^{2n}\sin^{2n}{\theta}d\theta\\ &=&\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_{n}}{n!}k^{2n}\int_{0}^{\frac{\pi}{2}}\sin^{2n}{\theta}d\theta\\ &=&\sum_{n=0}^{\infty}\frac{(\frac{1}{2})_{n}}{n!}k^{2n}\frac{(\frac{1}{2})_{n}}{n!}\frac{\pi}{2}\\ &=&\frac{\pi}{2}{}_{2}F_{1}(\frac{1}{2},\frac{1}{2};1;k^{2}) \end{eqnarray}

\begin{eqnarray} K(\frac{1}{\sqrt{2}})&=&\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1-\frac{1}{2}\sin^{2}{\theta}}}\\ &=&\sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{2-\sin^{2}{\theta}}}\quad(\sin{\theta}=x)\\ &=&\sqrt{2}\int_{0}^{1}\frac{dx}{\sqrt{(1-x^{2})(2-x^{2})}}\quad(x^2=y)\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{dy}{\sqrt{y(1-y)(2-y)}}\quad(1-y=z)\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{dz}{\sqrt{z(1-z)(1+z)}}\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{dz}{\sqrt{z(1-z^{2})}}\\ &=&\frac{1}{2\sqrt{2}}\int_{0}^{1}\frac{dw}{w^{\frac{3}{4}}(1-w)^{\frac{1}{2}}}\\ &=&\frac{1}{2\sqrt{2}}B(\frac{1}{4},\frac{1}{2})\\ &=&\frac{\Gamma(\frac{1}{4})\Gamma(\frac{1}{2})}{2\sqrt{2}\Gamma(\frac{3}{4})}\\ &=&\frac{\Gamma(\frac{1}{4})^{2}\sqrt{\pi}}{2\sqrt{2}\Gamma(\frac{1}{4})\Gamma(1-\frac{1}{4})}\\ &=&\frac{\Gamma(\frac{1}{4})^{2}}{4\sqrt{\pi}} \end{eqnarray}

\begin{eqnarray} K(i\frac{1}{\sqrt{2}})&=&\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{1+\frac{1}{2}\sin^{2}{\theta}}}\\ &=&\sqrt{2}\int_{0}^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{2+\sin^{2}{\theta}}}\quad(\sin{\theta}=x)\\ &=&\sqrt{2}\int_{0}^{1}\frac{dx}{\sqrt{(1-x^{2})(2+x^{2})}}\quad(x^{2}=y)\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{dy}{\sqrt{y(1-y)(2+y)}}\quad(y=1-z)\\ &=&\frac{1}{\sqrt{2}}\int_{0}^{1}\frac{dz}{\sqrt{z(1-z)(3-z)}}\\ &=&\frac{1}{\sqrt{6}}\int_{0}^{1}z^{\frac{1}{2}-1}(1-z)^{\frac{1}{2}-1}(1-\frac{1}{3}z)^{-\frac{1}{2}}dz\\ &=&\frac{1}{\sqrt{6}}\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(1)}{}_{2}F_{1}(\frac{1}{2},\frac{1}{2};1;\frac{1}{3})\\ &=&\frac{\pi}{\sqrt{6}}{}_{2}F_{1}(\frac{1}{2},\frac{1}{2};1;\frac{1}{3}) \end{eqnarray}

[1]
\begin{eqnarray} \frac{d^{n}}{dx^{n}}\log{(1-x)}&=&-\frac{(n-1)!}{(1-x)^{n}} \end{eqnarray}
[2]
\begin{eqnarray} -\log{(1-x)}&=&\sum_{n=1}^{\infty}\frac{1}{n}x^{n}\\ &=&\sum_{n=1}^{\infty}\frac{(n-1)!(n-1)!}{n!}\frac{x^{n}}{(n-1)!}\\ &=&x\sum_{n=0}^{\infty}\frac{n!n!}{(n+1)!}\frac{x^{n}}{n!}\\ &=&x{}_{2}F_{1}(1,1;2;x) \end{eqnarray}

\begin{eqnarray} \psi(\alpha,\beta,\gamma,z)&=&\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}(1-zx)^{-\gamma}dx\\ &=&\sum_{n=0}^{\infty}\frac{(\gamma)_{n}}{n!}z^{n}\int_{0}^{1}x^{\alpha+n-1}(1-x)^{\beta-1}dx\\ &=&\sum_{n=0}^{\infty}\frac{(\gamma)_{n}}{n!}z^{n}\frac{\Gamma(\alpha+n)\Gamma(\beta)}{\Gamma(\alpha+\beta+n)}\\ &=&\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\sum_{n=0}^{\infty}\frac{(\alpha)_{n}(\gamma)_{n}}{(\alpha+\beta)_{n}}\frac{z^{n}}{n!}\\ &=&\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}{}_{2}F_{1}(\alpha,\gamma;\alpha+\beta;z) \end{eqnarray}

\begin{eqnarray} (\frac{1}{N})_{n}(\frac{2}{N})_{n}\cdots(\frac{N-1}{N})_{n}&=&\frac{1}{N}\frac{2}{N}\cdots\frac{N-1}{N}\frac{N+1}{N}\frac{N+2}{N}\cdots\frac{2N-1}{N}\cdots\frac{(n-1)N+1}{N}\frac{(n-1)N+2}{N}\cdots\frac{nN-1}{N}\\ &=&\frac{(nN)!}{n!N^{nN}} \end{eqnarray}

[1]$R=1$とした場合。
\begin{eqnarray} (\frac{QN+1}{N})_{n}(\frac{QN+2}{N})_{n}\cdots(\frac{QN+N-1}{N})_{n}&=&\frac{QN+1}{N}\frac{QN+2}{N}\cdots\frac{QN+N-1}{N}\frac{QN+N+1}{N}\frac{QN+N+2}{N}\cdots\frac{QN+N+N-1}{N}\cdots\frac{(Q+n-1)N+1}{N}\frac{(Q+n-1)N+2}{N}\cdots\frac{(n+Q)N-1}{N}\\ &=&\frac{Q!\{(Q+n)N\}!}{(Q+n)!(QN)!N^{nN}} \end{eqnarray}
[2]$R\neq 1$とした場合。
\begin{eqnarray} \color{red}(\frac{QN+R}{N})_{n}(\frac{QN+R+1}{N})_{n}\cdots(\frac{QN+N-1}{N})_{n}\color{PineGreen}(\frac{(Q+1)N+1}{N})_{n}\cdots(\frac{(Q+1)N+R-1}{N})_{n}&=&\color{red}\frac{QN+R}{N}\frac{QN+R+1}{N}\cdots\frac{QN+N-1}{N}\color{PineGreen}\frac{(Q+1)N+1}{N}\frac{(Q+1)N+2}{N}\cdots\frac{(Q+1)N+R-1}{N}\color{red}\frac{(Q+1)N+R}{N}\frac{(Q+1)N+R+1}{N}\cdots\frac{(Q+1)N+N-1}{N}\color{PineGreen}\frac{(Q+2)N+1}{N}\cdots\frac{(Q+2)N+R-1}{N}\color{red}\frac{(Q+n-1)N+R}{N}\frac{(Q+n-1)N+R+1}{N}\cdots\frac{(Q+n-1)N+N-1}{N}\color{PineGreen}\frac{(Q+n)N+1}{N}\cdots\frac{(Q+n)N+R-1}{N}\\ &=&\frac{Q!\{(Q+n)N+R-1\}!}{(Q+n)!(QN+R-1)!N^{nN}}\\ \end{eqnarray}

(1)の条件により、$|a|\lt n$の様にとる。すると
\begin{equation} \exists C\in\mathbb{R}_{+}\ s.t.\ |f(t)|\leq Ce^{at}\lt Ce^{nt} \end{equation}
ゆえに
\begin{eqnarray} |\int_{0}^{\infty}f(x)e^{-nx}dx|&\leq&\int_{0}^{\infty}|f(x)|e^{-nx}dx\\ &\lt&C\int_{0}^{\infty}e^{ax}e^{-nx}dx\\ &=&C\int_{0}^{\infty}e^{-(n-a)x}dx\\ &=&\frac{C}{n-a}\lt+\infty \end{eqnarray}
絶対収束する。
次にその様な自然数$n$を取ります。
\begin{eqnarray} I_{n}&=&\int_{0}^{\infty}f(x)e^{-nx}dx\\ &\simeq&\int_{0}^{\infty}\sum_{m=0}^{\infty}c_{m}x^{a_{m}}e^{-nx}dx\quad(nx=y)\\ &=&\sum_{m=0}^{\infty}\frac{c_{m}}{n^{a_{m+1}}}\int_{0}^{\infty}x^{a_{m+1-1}}e^{-x}dx\\ &=&\sum_{m=0}^{\infty}c_{m}\frac{\Gamma(a_{m}+1)}{n^{a_{m+1}}} \end{eqnarray}

下の図を参照すればよい。

Abelの総和法

[1]右辺の積分について少し考察すれば分かる。
\begin{eqnarray} \int_{0}^{\lfloor x\rfloor}A(y)f^{'}(y)dy&=&\sum_{k=1}^{\lfloor x\rfloor}\int_{k-1}^{k}A(y)f^{'}(y)dy\\ &=&\sum_{k=1}^{\lfloor x\rfloor}\lim_{\epsilon\rightarrow0+0}\int_{k-1}^{k-\epsilon}A(y)f^{'}(y)dy\\ &=&\sum_{k=1}^{\lfloor x\rfloor}A(k-1)\lim_{\epsilon\rightarrow0+0}\int_{k-1}^{k-\epsilon}f^{'}(y)dy\\ &=&\sum_{k=1}^{\lfloor x\rfloor}A(k-1)\lim_{\epsilon\rightarrow0+0}\{f(k-\epsilon)-f(k-1)\}\\ &=&\sum_{k=1}^{\lfloor x\rfloor}A(k-1)\{f(k)-f(k-1)\} \end{eqnarray}
[2]$f(k)=b_{k}$とおくと、Abelの総和と一致する事が分かるので証明終了。

逆Laplace変換

解析的表示$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$が存在する場合のみ示す。
[1]仮定より、Laplace変換は常に定義出来なおかつ、それは絶対収束しますので次の様な計算が出来る。
\begin{eqnarray} \mathcal{L}[f](s)&=&\int_{0}^{\infty}f(z)e^{-sz}dz\\ &=&\int_{0}^{\infty}\sum_{n=0}^{\infty}a_{n}z^{n}e^{-sz}dz\\ &=&\sum_{n=0}^{\infty}a_{n}\int_{0}^{\infty}z^{n}e^{-sz}dz\\ &=&\sum_{n=0}^{\infty}\frac{a_{n}n!}{s^{n+1}} \end{eqnarray}
[2]この得られた級数について以下の計算を行います。
\begin{eqnarray} \mathcal{L}^{-1}[f](z)&=&\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\sum_{n=0}^{\infty}\frac{a_{n}n!}{s^{n+1}}e^{sz}ds+\int_{\Gamma}\sum_{n=0}^{\infty}\frac{a_{n}n!}{s^{n+1}}e^{sz}ds\\ &=&\sum_{n=0}^{\infty}\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{a_{n}n!}{s^{n+1}}\sum_{m=0}^{\infty}\frac{s^{m}z^{m}}{m!}ds\\ &=&\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{a_{n}n!z^{m}}{m!}\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{1}{s^{n-m+1}}ds\\ &=&\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{a_{n}n!z^{m}}{m!}\delta_{n,m}\\ &=&\sum_{n=0}^{\infty}a_{n}z^{n}\\ &=&f(z) \end{eqnarray}

[1]Abelの総和法により、任意の微分可能な関数$\varphi\in C^{1}(1,x)$に対して以下の式が成り立つ。
\begin{eqnarray} \left\{ \begin{array}{l} A(x)=\sum_{1\leq n\leq \lfloor x\rfloor}a_{n}\\ \sum_{1\leq n\leq \lfloor x\rfloor}a_{n}\varphi(n)=A(\lfloor x\rfloor)\varphi(\lfloor x\rfloor)-\int_{1}^{\lfloor x\rfloor}A(t)\varphi^{'}(t)dt \end{array} \right. \end{eqnarray}
[2]$\varphi(x)\coloneqq x^{-s}$とおく。すると下記の様に書き直せる。
\begin{eqnarray} \sum_{1\leq n\leq \lfloor x\rfloor}\frac{a_{n}}{n^{s}}=\frac{A(\lfloor x\rfloor)}{\lfloor x \rfloor^{s}}+s\int_{1}^{\lfloor x\rfloor}\frac{A(t)}{t^{s+1}}dt \end{eqnarray}
[3]$x\rightarrow\infty$とする。すると以下の式が成り立つ。
\begin{eqnarray} f(s)&=&s\int_{1}^{\infty}\frac{A(t)}{t^{s+1}}dt\quad(t=e^{\tau})\\ &=&s\int_{0}^{\infty}A(\tau)e^{-s\tau}d\tau \end{eqnarray}
[4]ゆえに
\begin{equation} \frac{f(s)}{s}=\int_{0}^{\infty}A(\tau)e^{-s\tau}d\tau\equiv\mathcal{L}[A(\tau)](s) \end{equation}
[5][4]の右辺はLaplace変換なのでLaplace変換の逆変換を行う事で
\begin{equation} A(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)}{s}e^{sx}ds \end{equation}
最後に$x=\log{y}$の様に置き換えて
\begin{equation} A(\log{y})=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)}{s}y^{s}ds \end{equation}