階乗冪(下降階乗・上昇階乗)の和分

by nomura · 2020年6月13日

(1)

\(m\in\mathbb{N}_{0}\cup\{-1\}\;\land\;n\in\mathbb{Z}\setminus\{-1\}\)
\[ \sum_{k=1}^{m}P(k,n)=\frac{1}{n+1}P(m+1,n+1) \]

(2)

\(m-n\in\mathbb{N}_{0}\cup\{-1\}\;\land\;n\in\mathbb{Z}\setminus\mathbb{N}\)
\[ \sum_{k=1}^{m}P^{-1}(k,n)=\frac{1}{n-1}\left((1-n)!-P(m-n+1,1-n)\right) \]

(3)

\(m+n\in\mathbb{N}_{0}\;\land\;n\in\mathbb{Z}\setminus\{-1\}\)
\[ \sum_{k=1}^{m}Q(k,n)=\frac{1}{n+1}Q(m,n+1) \]

(4)

\(m\in\mathbb{N}_{0}\;\land\;n\in\mathbb{Z}\setminus\{1\}\)
\[ \sum_{k=1}^{m}Q^{-1}(k,n)=\frac{1}{n-1}\left(\frac{1}{(n-1)!}-\frac{1}{Q(m+1,n-1)}\right) \]

(1)

\begin{align*} \sum_{k=0}^{m}P(k,n) & =\sum_{k=0}^{m}\left\{ \frac{1}{n+1}\left(P(k+1,n+1)-P(k,n+1)\right)\right\} \\ & =\frac{1}{n+1}\left(P(m+1,n+1)-P(0,n+1)\right)\\ & =\frac{1}{n+1}P(m+1,n+1) \end{align*}

(2)

\begin{align*} \sum_{k=1}^{m}P^{-1}(k,n) & =\sum_{k=1}^{m}\frac{1}{n-1}\left(\frac{1}{P(k-1,n-1)}-\frac{1}{P(k,n-1)}\right)\\ & =\frac{1}{n-1}\left(\frac{1}{P(0,n-1)}-\frac{1}{P(m,n-1)}\right)\\ & =\frac{1}{n-1}\left((1-n)!-P(m-n+1,1-n)\right) \end{align*}

(3)

\begin{align*} \sum_{k=1}^{m}Q(k,n) & =\sum_{k=1}^{m}\frac{1}{n+1}\left(Q(k,n+1)-Q(k-1,n+1)\right)\\ & =\frac{1}{n+1}\left(Q(m,n+1)-Q(0,n+1)\right)\\ & =\frac{1}{n+1}Q(m,n+1) \end{align*}

(4)

\begin{align*} \sum_{k=1}^{m}Q^{-1}(k,n) & =\sum_{k=1}^{m}\frac{1}{n-1}\left(\frac{1}{Q(k,n-1)}-\frac{1}{Q(k+1,n-1)}\right)\\ & =\frac{1}{n-1}\left(\frac{1}{Q(1,n-1)}-\frac{1}{Q(m+1,n-1)}\right)\\ & =\frac{1}{n-1}\left(\frac{1}{(n-1)!}-\frac{1}{Q(m+1,n-1)}\right) \end{align*}