(拡張)多重階乗の逆数和
(拡張)多重階乗の逆数和
\(n\in\mathbb{N}\)とする。
\(n\in\mathbb{N}\)とする。
(1)
\[ \sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right) \](2)
\[ \sum_{k=0}^{\infty}\frac{1}{\left(ak+b\right)!_{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(1-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right) \](3)
\[ \sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!^{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!^{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right) \](4)
\[ \sum_{k=0}^{\infty}\frac{1}{\left(ak+b\right)!^{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!^{a}}\left(1-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right) \]-
\(\Gamma\left(x\right)\)はガンマ関数、\(\Gamma\left(k,x\right)\)は第2種不完全ガンマ関数、\(n!_{p}\)は多重階乗、\(n!^{p}\)は拡張多重階乗(1)
\begin{align*} \sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}} & =\sum_{k=0}^{n}\frac{1}{m^{k}b!_{a}}\frac{\Gamma\left(\frac{b}{a}+1\right)}{\left(k+\frac{b}{a}\right)!}\\ & =\frac{\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\sum_{k=0}^{n}\frac{1}{a^{k}\left(k+\frac{b}{a}\right)!}\\ & =\frac{\Gamma\left(\frac{b}{a}+1\right)a^{\frac{b}{a}}}{b!_{a}}\sum_{k=0}^{n}\frac{1}{\left(k+\frac{b}{a}\right)!}\left(\frac{1}{a}\right)^{k+\frac{b}{a}}\\ & =\frac{\Gamma\left(\frac{b}{a}+1\right)a^{\frac{b}{a}}}{b!_{a}}\sum_{k=0}^{n}e^{\frac{1}{a}}\left(\frac{\Gamma\left(k+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(k+\frac{b}{a}+1\right)}-\frac{\Gamma\left(k+\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(k+\frac{b}{a}\right)}\right)\\ & =\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right) \end{align*}(2)
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{\left(ak+b\right)!_{a}} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}}\\ & =\lim_{n\rightarrow\infty}\frac{e^{\frac{1}{a}}\Gamma\left(\frac{b}{a}+1\right)a^{\frac{b}{a}}}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right)\\ & =\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(1-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right) \end{align*}(3)
(1)と同じ(4)
(2)と同じページ情報
| タイトル | (拡張)多重階乗の逆数和 |
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多重階乗同士の関係
\[
\left(qn+r\right)!^{n}=r!^{n}\frac{\left(qn+r\right)!_{n}}{r!_{n}}
\]
2重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(2k\right)!!}=\sqrt{e}\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)}
\]
拡張多重階乗の漸化式
\[
x!^{n}=x\left(x-n\right)!^{n}
\]
負の多重階乗
\[
\left(-\left(qn+r\right)\right)!_{n}=\frac{\left(-1\right)^{q}}{\left(qn-\left(n-r\right)\right)!_{n}}
\]