整数と半整数の逆数和
整数と半整数の逆数和
(1)
\[ \sum_{k=0}^{n}\frac{1}{k!}=e\frac{\Gamma(n+1,1)}{\Gamma\left(n+1\right)} \](2)
\[ \sum_{k=0}^{\infty}\frac{1}{k!}=e \](3)
\[ \sum_{k=0}^{n}\frac{1}{\left(k+\frac{1}{2}\right)!}=e\left(\frac{\Gamma\left(n+\frac{3}{2},1\right)}{\Gamma\left(n+\frac{3}{2}\right)}+\erf(1)-1\right) \](4)
\[ \sum_{k=0}^{\infty}\frac{1}{\left(k+\frac{1}{2}\right)!}=e\erf(1) \]-
\(\Gamma(x,y)\)は第2種不完全ガンマ関数、\(\erf(x)\)は誤差関数(1)
\begin{align*} \sum_{k=0}^{n}\frac{1}{k!} & =\sum_{k=0}^{n}\frac{1}{k!}1^{k}\\ & =\sum_{k=0}^{n}e^{x}\left(\frac{\Gamma\left(k+1,1\right)}{\Gamma\left(k+1\right)}-\frac{\Gamma\left(k,1\right)}{\Gamma\left(k\right)}\right)\\ & =e\frac{\Gamma(n+1,1)}{\Gamma\left(n+1\right)} \end{align*}(2)
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{k!} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{k!}\\ & =\lim_{n\rightarrow\infty}e\frac{\Gamma(n+1,1)}{\Gamma\left(n+1\right)}\\ & =e \end{align*}(2)-2
\[ \sum_{k=0}^{\infty}\frac{1}{k!}=e \](3)
\begin{align*} \sum_{k=0}^{n}\frac{1}{\left(k+\frac{1}{2}\right)!} & =\sum_{k=0}^{n}\frac{1}{\left(k+\frac{1}{2}\right)!}1^{k+\frac{1}{2}}\\ & =\sum_{k=0}^{n}e\left(\frac{\Gamma\left(k+\frac{1}{2}+1,1\right)}{\Gamma\left(k+\frac{1}{2}+1\right)}-\frac{\Gamma\left(k+\frac{1}{2},1\right)}{\Gamma\left(k+\frac{1}{2}\right)}\right)\\ & =e\left(\frac{\Gamma\left(n+\frac{3}{2},1\right)}{\Gamma\left(n+\frac{3}{2}\right)}-\frac{\Gamma\left(\frac{1}{2},1\right)}{\Gamma\left(\frac{1}{2}\right)}\right)\\ & =e\left(\frac{\Gamma\left(n+\frac{3}{2},1\right)}{\Gamma\left(n+\frac{3}{2}\right)}+\frac{\gamma\left(\frac{1}{2},1\right)}{\Gamma\left(\frac{1}{2}\right)}-1\right)\\ & =e\left(\frac{\Gamma\left(n+\frac{3}{2},1\right)}{\Gamma\left(n+\frac{3}{2}\right)}+\erf(1)-1\right) \end{align*}(4)
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{\left(k+\frac{1}{2}\right)!} & =\lim_{n\rightarrow\infty}\sum_{k=0}^{n}\frac{1}{\left(k+\frac{1}{2}\right)!}\\ & =\lim_{n\rightarrow\infty}e\left(\frac{\Gamma\left(n+\frac{3}{2},1\right)}{\Gamma\left(n+\frac{3}{2}\right)}+\erf(1)-1\right)\\ & =e\erf(1) \end{align*}(4)-2
\begin{align*} \sum_{k=0}^{\infty}\frac{1}{\left(k+\frac{1}{2}\right)!} & =\sum_{k=0}^{\infty}\frac{1}{\left(k+\frac{1}{2}\right)!}1^{k+\frac{1}{2}}\\ & =e\left(1-\frac{\Gamma\left(\frac{1}{2},x\right)}{\Gamma\left(\frac{1}{2}\right)}\right)\\ & =e\left(\frac{\gamma\left(\frac{1}{2},x\right)}{\Gamma\left(\frac{1}{2}\right)}\right)\\ & =e\erf(1) \end{align*}ページ情報
タイトル | 整数と半整数の逆数和 |
URL | https://www.nomuramath.com/dd4q7y7s/ |
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2重和の変換
\[
\sum_{m=a}^{\infty}\sum_{n=b}^{\infty}f(m,n)=\sum_{t=a+b}^{\infty}\sum_{s=a}^{t-b}f(s,t-s)
\]
2重根号の逆数の総和
\[
\sum_{k=1}^{n}\frac{1}{\sqrt{k+\sqrt{k^{2}-1}}}=\frac{\sqrt{2}}{2}\left(\sqrt{n+1}+\sqrt{n}-1\right)
\]
1次式の逆n乗和
\[
\sum_{k=1}^{m}\frac{1}{\left(\alpha k+\beta\right)^{n}}=\frac{\left(-1\right)^{n-1}}{\alpha^{n}\left(n-1\right)!}\left\{ \psi^{\left(n-1\right)}\left(m+1+\frac{\beta}{\alpha}\right)-\psi^{\left(n-1\right)}\left(1+\frac{\beta}{\alpha}\right)\right\}
\]