2重階乗の逆数和
2重階乗の逆数和
(1)
\[ \sum_{k=0}^{n}\frac{1}{\left(2k\right)!!}=\sqrt{e}\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)} \](2)
\[ \sum_{k=0}^{\infty}\frac{1}{\left(2k\right)!!}=\sqrt{e} \](3)
\[ \sum_{k=0}^{n}\frac{1}{\left(2k+1\right)!!}=\sqrt{\frac{e\pi}{2}}\left(\frac{\Gamma\left(n+\frac{3}{2},\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}-1+\erf\left(\frac{\sqrt{2}}{2}\right)\right) \](4)
\[ \sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)!!}=\sqrt{\frac{e\pi}{2}}\erf\left(\frac{\sqrt{2}}{2}\right) \]-
\(\Gamma\left(x\right)\)はガンマ関数、\(\Gamma\left(k,x\right)\)は第2種不完全ガンマ関数、\(n!!\)は2重階乗、\(\erf\left(x\right)\)は誤差関数(1)
\begin{align*} \sum_{k=0}^{n}\frac{1}{\left(2k\right)!!} & =\sum_{k=0}^{n}\frac{1}{2^{k}k!}\\ & =\sum_{k=0}^{n}\frac{1}{k!}\left(\frac{1}{2}\right)^{k}\\ & =\sum_{k=+0}^{n}\sqrt{e}\left(\frac{\Gamma\left(k+1,\frac{1}{2}\right)}{\Gamma\left(k+1\right)}-\frac{\Gamma\left(k,\frac{1}{2}\right)}{\Gamma\left(k\right)}\right)\\ & =\sqrt{e}\left(\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)}-\lim_{k\rightarrow0}\frac{\Gamma\left(k,\frac{1}{2}\right)}{\Gamma\left(k\right)}\right)\\ & =\sqrt{e}\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)} \end{align*}(2)
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{\left(2k\right)!!} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{\left(2k\right)!!}\\ & =\lim_{n\rightarrow\infty}\sqrt{e}\frac{\Gamma\left(n+1,\frac{1}{2}\right)}{\Gamma\left(n+1\right)}\\ & =\sqrt{e} \end{align*}(2)-2
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{\left(2k\right)!!} & =\sum_{k=0}^{\infty}\frac{1}{2^{k}k!}\\ & =\sum_{k=0}^{\infty}\frac{1}{k!}\left(\frac{1}{2}\right)^{k}\\ & =\sqrt{e} \end{align*}(3)
\begin{align*} \sum_{k=0}^{n}\frac{1}{\left(2k+1\right)!!} & =\sqrt{\pi}\sum_{k=0}^{n}\frac{1}{2^{k+1}\left(k+\frac{1}{2}\right)!}\\ & =\sqrt{\frac{\pi}{2}}\sum_{k=0}^{n}\frac{1}{\left(k+\frac{1}{2}\right)!}\left(\frac{1}{2}\right)^{k+\frac{1}{2}}\\ & =\sqrt{\frac{\pi}{2}}\sum_{k=0}^{n}\sqrt{e}\left(\frac{\Gamma\left(k+\frac{1}{2}+1,\frac{1}{2}\right)}{\Gamma\left(k+\frac{1}{2}+1\right)}-\frac{\Gamma\left(k+\frac{1}{2},\frac{1}{2}\right)}{\Gamma\left(k+\frac{1}{2}\right)}\right)\\ & =\sqrt{\frac{e\pi}{2}}\left(\frac{\Gamma\left(n+\frac{3}{2},\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}-\frac{\Gamma\left(\frac{1}{2},\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}\right)\\ & =\sqrt{\frac{e\pi}{2}}\left(\frac{\Gamma\left(n+\frac{3}{2},\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}-\frac{\Gamma\left(\frac{1}{2}\right)-\gamma\left(\frac{1}{2},\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}\right)\\ & =\sqrt{\frac{e\pi}{2}}\left(\frac{\Gamma\left(n+\frac{3}{2},\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}-\frac{\Gamma\left(\frac{1}{2}\right)-\sqrt{\pi}\erf\left(\sqrt{\frac{1}{2}}\right)}{\Gamma\left(\frac{1}{2}\right)}\right)\\ & =\sqrt{\frac{e\pi}{2}}\left(\frac{\Gamma\left(n+\frac{3}{2},\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}-1+\erf\left(\frac{\sqrt{2}}{2}\right)\right) \end{align*}(4)
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{\left(2k+1\right)!!} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{\left(2k+1\right)!!}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{e\pi}{2}}\left(\frac{\Gamma\left(n+\frac{3}{2},\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{2}\right)}-1+\erf\left(\frac{\sqrt{2}}{2}\right)\right)\\ & =\sqrt{\frac{e\pi}{2}}\erf\left(\frac{\sqrt{2}}{2}\right) \end{align*}ページ情報
| タイトル | 2重階乗の逆数和 |
| URL | https://www.nomuramath.com/d0t3pqsc/ |
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階乗の多重階乗表示
\[
n!=\prod_{k=0}^{j-1}\left(n-k\right)!_{j}
\]
(拡張)多重階乗と階乗の関係
\[
\left(an+b\right)!_{a}=\frac{a^{n}b!_{a}\left(n+\frac{b}{a}\right)!}{\left(\frac{b}{a}\right)!}
\]
拡張多重階乗の簡単な値
\[
0!^{n}=\frac{1}{\sqrt[n]{n}\left(\frac{1}{n}\right)!}
\]
(拡張)多重階乗の逆数和
\[
\sum_{k=0}^{n}\frac{1}{\left(ak+b\right)!_{a}}=\frac{e^{\frac{1}{a}}a^{\frac{b}{a}}\Gamma\left(\frac{b}{a}+1\right)}{b!_{a}}\left(\frac{\Gamma\left(n+\frac{b}{a}+1,\frac{1}{a}\right)}{\Gamma\left(n+\frac{b}{a}+1\right)}-\frac{\Gamma\left(\frac{b}{a},\frac{1}{a}\right)}{\Gamma\left(\frac{b}{a}\right)}\right)
\]