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。
\[ \sum_{k=1}^{m}\frac{1}{\left(\alpha k+\beta\right)^{n}}=\frac{\left(-1\right)^{n-1}}{\alpha^{n}\left(n-1\right)!}\left\{ \psi^{\left(n-1\right)}\left(m+1+\frac{\beta}{\alpha}\right)-\psi^{\left(n-1\right)}\left(1+\frac{\beta}{\alpha}\right)\right\} \]
\(\alpha\ne0\;\land\;n\in\mathbb{N}\)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 |
2重根号の逆数の総和
\[
\sum_{k=1}^{n}\frac{1}{\sqrt{k+\sqrt{k^{2}-1}}}=\frac{\sqrt{2}}{2}\left(\sqrt{n+1}+\sqrt{n}-1\right)
\]
2重和の変換
\[
\sum_{m=a}^{\infty}\sum_{n=b}^{\infty}f(m,n)=\sum_{t=a+b}^{\infty}\sum_{s=a}^{t-b}f(s,t-s)
\]
整数と半整数の逆数和
\[
\sum_{k=0}^{n}\frac{1}{k!}=e\frac{\Gamma(n+1,1)}{\Gamma\left(n+1\right)}
\]