1次式の逆n乗和

1次式の逆n乗和

\(\alpha\ne0\;\land\;n\in\mathbb{N}\)とする。

\[ \sum_{k=1}^{m}\frac{1}{\left(\alpha k+\beta\right)^{n}}=\frac{\left(-1\right)^{n-1}}{\alpha^{n}\left(n-1\right)!}\left\{ \psi^{\left(n-1\right)}\left(m+1+\frac{\beta}{\alpha}\right)-\psi^{\left(n-1\right)}\left(1+\frac{\beta}{\alpha}\right)\right\} \]

-

\(\psi^{\left(n\right)}\left(z\right)\)はポリガンマ関数

\begin{align*} \sum_{k=1}^{m}\frac{1}{\left(\alpha k+\beta\right)^{n}} & =\frac{1}{\alpha^{n}}\sum_{k=1}^{m}\frac{1}{\left(k+\frac{\beta}{\alpha}\right)^{n}}\\ & =\frac{1}{\alpha^{n}}\frac{1}{P\left(-1,n-1\right)}\sum_{k=1}^{m}\left[\frac{d^{n-1}}{dz^{n-1}}\frac{1}{z}\right]_{z=k+\frac{\beta}{\alpha}}\\ & =\frac{\left(-1\right)^{n-1}}{\alpha^{n}\left(n-1\right)!}\sum_{k=1}^{m}\left[\frac{d^{n-1}}{dz^{n-1}}\left\{ \psi\left(z+1\right)-\psi\left(z\right)\right\} \right]_{z=k+\frac{\beta}{\alpha}}\\ & =\frac{\left(-1\right)^{n-1}}{\alpha^{n}\left(n-1\right)!}\sum_{k=1}^{m}\left\{ \psi^{\left(n-1\right)}\left(k+\frac{\beta}{\alpha}+1\right)-\psi^{\left(n-1\right)}\left(k+\frac{\beta}{\alpha}\right)\right\} \\ & =\frac{\left(-1\right)^{n-1}}{\alpha^{n}\left(n-1\right)!}\left\{ \psi^{\left(n-1\right)}\left(m+1+\frac{\beta}{\alpha}\right)-\psi^{\left(n-1\right)}\left(1+\frac{\beta}{\alpha}\right)\right\} \end{align*}

ページ情報

タイトル

1次式の逆n乗和

URL

https://www.nomuramath.com/vog7el54/

SNSボタン