多重階乗

(拡張)多重階乗の逆数和

by nomura · 2021年3月17日

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(拡張)多重階乗の逆数和

n \in N

とする。

(1)

\sum_{k = 0}^{n} \frac{1}{(a k + b)!_{a}} = \frac{e^{\frac{1}{a}} a^{\frac{b}{a}} Γ (\frac{b}{a} + 1)}{b!_{a}} (\frac{Γ (n + \frac{b}{a} + 1, \frac{1}{a})}{Γ (n + \frac{b}{a} + 1)} - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})})

(2)

\sum_{k = 0}^{\infty} \frac{1}{(a k + b)!_{a}} = \frac{e^{\frac{1}{a}} a^{\frac{b}{a}} Γ (\frac{b}{a} + 1)}{b!_{a}} (1 - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})})

(3)

\sum_{k = 0}^{n} \frac{1}{(a k + b)!^{a}} = \frac{e^{\frac{1}{a}} a^{\frac{b}{a}} Γ (\frac{b}{a} + 1)}{b!^{a}} (\frac{Γ (n + \frac{b}{a} + 1, \frac{1}{a})}{Γ (n + \frac{b}{a} + 1)} - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})})

(4)

\sum_{k = 0}^{\infty} \frac{1}{(a k + b)!^{a}} = \frac{e^{\frac{1}{a}} a^{\frac{b}{a}} Γ (\frac{b}{a} + 1)}{b!^{a}} (1 - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})})

-

Γ (x)

はガンマ関数、

Γ (k, x)

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{aligned} \sum_{k = 0}^{n} \frac{1}{(a k + b)!_{a}} & = \sum_{k = 0}^{n} \frac{1}{m^{k} b!_{a}} \frac{Γ (\frac{b}{a} + 1)}{(k + \frac{b}{a})!} \\ = \frac{Γ (\frac{b}{a} + 1)}{b!_{a}} \sum_{k = 0}^{n} \frac{1}{a^{k} (k + \frac{b}{a})!} \\ = \frac{Γ (\frac{b}{a} + 1) a^{\frac{b}{a}}}{b!_{a}} \sum_{k = 0}^{n} \frac{1}{(k + \frac{b}{a})!} {(\frac{1}{a})}^{k + \frac{b}{a}} \\ = \frac{Γ (\frac{b}{a} + 1) a^{\frac{b}{a}}}{b!_{a}} \sum_{k = 0}^{n} e^{\frac{1}{a}} (\frac{Γ (k + \frac{b}{a} + 1, \frac{1}{a})}{Γ (k + \frac{b}{a} + 1)} - \frac{Γ (k + \frac{b}{a}, \frac{1}{a})}{Γ (k + \frac{b}{a})}) \\ = \frac{e^{\frac{1}{a}} a^{\frac{b}{a}} Γ (\frac{b}{a} + 1)}{b!_{a}} (\frac{Γ (n + \frac{b}{a} + 1, \frac{1}{a})}{Γ (n + \frac{b}{a} + 1)} - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})}) \end{aligned}

(2)

\begin{aligned} \sum_{k = 0}^{\infty} \frac{1}{(a k + b)!_{a}} & = lim_{n \to \infty} \sum_{k = 0}^{n} \frac{1}{(a k + b)!_{a}} \\ = lim_{n \to \infty} \frac{e^{\frac{1}{a}} Γ (\frac{b}{a} + 1) a^{\frac{b}{a}}}{b!_{a}} (\frac{Γ (n + \frac{b}{a} + 1, \frac{1}{a})}{Γ (n + \frac{b}{a} + 1)} - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})}) \\ = \frac{e^{\frac{1}{a}} a^{\frac{b}{a}} Γ (\frac{b}{a} + 1)}{b!_{a}} (1 - \frac{Γ (\frac{b}{a}, \frac{1}{a})}{Γ (\frac{b}{a})}) \end{aligned}

(3)

(1)と同じ

(4)

(2)と同じ

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(拡張)多重階乗と階乗の関係

(a n + b)!_{a} = \frac{a^{n} b!_{a} (n + \frac{b}{a})!}{(\frac{b}{a})!}

拡張多重階乗の簡単な値

0!^{n} = \frac{1}{\sqrt[n]{n} (\frac{1}{n})!}

多重階乗同士の関係

(q n + r)!^{n} = r!^{n} \frac{(q n + r)!_{n}}{r!_{n}}

階乗の多重階乗表示

n! = \prod_{k = 0}^{j - 1} (n - k)!_{j}