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*}

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ガンマ関数を含む極限

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