解析入門Ⅰ 第Ⅳ章積分法 §５一変数函数の積分　大問4 問題の解法

間違いがあればコメントで教えていただければ幸いです。

$${\bf{4})}$$次の一般項を持つ数列の極限を求めよ。

$${(\mathrm{i})\quad a_{n}\!=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}}$$

$$
\displaystyle\begin{aligned}a_{n}\!&=\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}\\&=\sum^{n}_{k=1}\dfrac{1}{n+k}=\dfrac{1}{n}\sum^{n}_{k=1}\dfrac{1}{1+\left(\dfrac{k}{n}\right)}\\数&列の極限を取ると\\\lim_{n\to\infty}a_{n}&=\lim_{n\to\infty}\dfrac{1}{n}\sum^{n}_{k=1}\dfrac{1}{1+\left(\dfrac{k}{n}\right)}=\int^{1}_{0}\dfrac{1}{1+x}dx\\&=\Big[\log|1+x|\Big]^{1}_{0}=\log{2}\end{aligned}
$$

$${(\mathrm{ii})\quad a_{n}\!=\dfrac{1}{\sqrt{n^{2}+n}}+\dfrac{1}{\sqrt{n^{2}+2n}}+\cdots+\dfrac{1}{\sqrt{n^{2}+n^{2}}}}$$

$$
\displaystyle\begin{aligned}a_{n}\!&=\dfrac{1}{\sqrt{n^{2}+n}}+\dfrac{1}{\sqrt{n^{2}+2n}}+\cdots+\dfrac{1}{\sqrt{n^{2}+n^{2}}}\\&=\sum^{n}_{k=1}\dfrac{1}{\sqrt{n^{2}+kn}}=\dfrac{1}{n}\sum^{n}_{k=1}\dfrac{1}{\sqrt{1+\left(\dfrac{k}{n}\right)}}\\数&列の極限を取ると\\\lim_{n\to\infty}a_{n}&=\lim_{n\to\infty}\dfrac{1}{n}\sum^{n}_{k=1}\dfrac{1}{\sqrt{1+\left(\dfrac{k}{n}\right)}}=\int^{1}_{0}\!\dfrac{1}{\sqrt{1+x}}\:dx\\&=\Big[2\sqrt{1+x}\Big]^{1}_{0}=2\sqrt{2}-2\end{aligned}
$$

$${(\mathrm{iii})\quad a_{n}\!=\dfrac{\sqrt[n]{n!}}{n}}$$

$$
\displaystyle\begin{aligned}a_{n}\!&=\dfrac{\sqrt[n]{n!}}{n}=\sqrt[n]{\dfrac{n(n-1)\cdots1}{n^{n}}}\\&=\sqrt[n]{\left(1-\dfrac{0}{n}\right)\left(1-\dfrac{1}{n}\right)\cdots\left(1-\dfrac{n-1}{n}\right)}\\数&列の自然対数を取ると\\\ln{a_{n}}&=\ln\sqrt[n]{\left(1-\dfrac{0}{n}\right)\left(1-\dfrac{1}{n}\right)\cdots\left(1-\dfrac{n-1}{n}\right)}\\&=\dfrac{1}{n}\ln{\left(1-\dfrac{0}{n}\right)\left(1-\dfrac{1}{n}\right)\cdots\left(1-\dfrac{n-1}{n}\right)}\\&=\dfrac{1}{n}\left[\ln{\left(1-\dfrac{0}{n}\right)}+\cdots\ln{\left(1-\dfrac{n-1}{n}\right)}\right]\\&=\dfrac{1}{n}\sum^{n-1}_{k=0}\ln{\left(1-\dfrac{k}{n}\right)}\end{aligned}
$$

$$
\displaystyle\begin{aligned}自&然対数の極限を取ると\\\lim_{n\to\infty}\ln{a_{n}}&=\lim_{n\to\infty}\dfrac{1}{n}\sum^{n-1}_{k=0}\ln{\left(1-\dfrac{k}{n}\right)}\\&=\int^{1}_{0}\ln{(1-x)}dx\\&=\int^{1}_{0}-(1-x)^{\prime}\ln{(1-x)}dx\\&=\Big[\!-\!(1-x)\ln{(1-x)}\Big]^{1}_{0}+\int^{1}_{0}(1-x)\dfrac{-1}{1-x}dx\\&=-\!\!\lim_{\alpha\to1-0}(1-\alpha)\ln{(1-\alpha)}-1\end{aligned}
$$

$$
\displaystyle\lim_{\alpha\to1-0}(1-\alpha)\ln{(1-\alpha)}=\lim_{\alpha\to1-0}\dfrac{\ln{(1-\alpha)}}{(1-\alpha)^{-1}}\\ロピタルの定理より、\\\begin{aligned}\lim_{\alpha\to1-0}\dfrac{\ln{(1-\alpha)}}{(1-\alpha)^{-1}}&=\lim_{\alpha\to1-0}\dfrac{\dfrac{-1}{1-\alpha}}{-(1-\alpha)^{-2}}=\lim_{\alpha\to1-0}(1-\alpha)\\&=0\end{aligned}
$$

$$
\displaystyle\therefore\lim_{n\to\infty}\ln{a_{n}}=-1\\よって\lim_{n\to\infty}a_{n}=\dfrac{1}{e}
$$