ガンマ関数を含む極限
ガンマ関数を含む極限値
(1)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}=1 \](2)
\[ \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}=\sqrt{2} \](1)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)}\sqrt{n+\frac{1}{2}}\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(n+1\right)}}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{\sqrt{n}\sqrt{n+\frac{1}{2}}}{n}}\\ & =\lim_{n\rightarrow\infty}\left(1+\frac{1}{2n}\right)^{\frac{1}{4}}\\ & =1 \end{align*}(1)-2
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(n\right)}{\Gamma\left(n+\frac{1}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\frac{n+1}{2}}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\\ & =\frac{1}{\sqrt{2}}\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}}{\sqrt{n}}\sqrt{n}\frac{2}{\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\sqrt{\frac{2}{\pi}}\lim_{n\rightarrow\infty}\frac{\sqrt{n+1}}{\sqrt{n}}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\sqrt{\frac{2}{\pi}}\sqrt{\frac{\pi}{2}}\\ & =1 \end{align*}(2)
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} & =\lim_{n\rightarrow\infty}\sqrt{\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)}\sqrt{n+1}\frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}}\\ & =\lim_{n\rightarrow\infty}\sqrt{\frac{2\sqrt{n}\sqrt{n+1}}{n+1}}\\ & =\sqrt{2}\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{-\frac{1}{4}}\\ & =\sqrt{2} \end{align*}(2)-2
\begin{align*} \lim_{n\rightarrow\infty}\sqrt{n}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n+2}{2}\right)} & =\frac{2}{\Gamma\left(\frac{1}{2}\right)}\lim_{n\rightarrow\infty}\sqrt{n}\int_{0}^{\frac{\pi}{2}}\sin^{n}\theta d\theta\\ & =\frac{2}{\sqrt{\pi}}\sqrt{\frac{\pi}{2}}\\ & =\sqrt{2} \end{align*}ページ情報
| タイトル | ガンマ関数を含む極限 |
| URL | https://www.nomuramath.com/hkc3vxq1/ |
| SNSボタン |
ディガンマ関数・ポリガンマ関数の級数表示・テイラー展開と調和数・一般化調和数
\[
\psi\left(z\right)=-\gamma+H_{z-1}
\]
1次式の総乗と階乗
\[
\prod_{k=a}^{b}\left(kn+r\right)=n^{b-a+1}\frac{\left(b+\frac{r}{n}\right)!}{\Gamma\left(a+\frac{r}{n}\right)}
\]
ディガンマ関数・ポリガンマ関数の相反公式
\[
\psi\left(1-z\right)-\psi\left(z\right)=\pi\tan^{-1}\left(\pi z\right)
\]
偶数と奇数の2重階乗
\[
\left(2n+1\right)!!=2^{n+1}\frac{\left(n+\frac{1}{2}\right)!}{\Gamma\left(\frac{1}{2}\right)}
\]