第4題和第9題
乍看之下好像都是黎曼和，想說是不是弄錯了，仔細算了一下才發現不一樣。
第4題
\(\displaystyle \lim_{n\to \infty}\left(\frac{1}{\sqrt{n^2+2n}}+\frac{1}{\sqrt{n^2+4n}}+\ldots+\frac{1}{\sqrt{n^2+2n^2}}\right)=\)
[解答]
(黎曼和)
\( \displaystyle \lim\limits_{n \to \infty} (\frac{1}{\sqrt{n^2 +2n}} + \frac{1}{\sqrt{n^2 +4n}} + ... + \frac{1}{\sqrt{n^2 +2n^2}} = \frac{1}{2} \lim\limits_{n \to \infty} \frac{2}{n} ( \frac{1}{\sqrt{1 +\frac{2}{n}}} + \frac{1}{\sqrt{1 +\frac{4}{n}}} + ... + \frac{1}{\sqrt{1 +\frac{2n}{n}}} = 0.5 \int_0^2 \frac{1}{\sqrt{1+x}} \mathrm{d} x =0.5*2 (\sqrt{1+x} |^2_0) = \sqrt{3} -1 ) \)
第9題
\(\displaystyle \lim_{n\to \infty}\left(\frac{1}{\sqrt{3n^2+1}}+\frac{1}{\sqrt{3n^2+2}}+\ldots+\frac{1}{\sqrt{3n^2+2n}}\right)=\)
[解答]
(夾擠定理)
\( \displaystyle \lim\limits_{n \to \infty} \frac{2n}{\sqrt{3n^2 }} = \lim\limits_{n \to \infty} (\frac{1}{\sqrt{3n^2 }} + \frac{1}{\sqrt{3n^2 }} + ... + \frac{1}{\sqrt{3n^2}} ) > \lim\limits_{n \to \infty} (\frac{1}{\sqrt{3n^2 + 1}} + \frac{1}{\sqrt{3n^2 + 2}} + ... + \frac{1}{\sqrt{3n^2 + 2n}} ) > \lim\limits_{n \to \infty} (\frac{1}{\sqrt{3n^2 +2n }} + \frac{1}{\sqrt{3n^2 +2n }} + ... + \frac{1}{\sqrt{3n^2+2n}}) = \lim\limits_{n \to \infty} \frac{2n}{\sqrt{3n^2 +2n}} \)
故 \( \displaystyle \lim\limits_{n \to \infty} \frac{2n}{\sqrt{3n^2 }} > 原式 > \lim\limits_{n \to \infty} \frac{2n}{\sqrt{3n^2 +2n}} \)，又\( \displaystyle \lim\limits_{n \to \infty} \frac{2n}{\sqrt{3n^2 }} = \frac{2}{\sqrt{3}}、 \lim\limits_{n \to \infty} \frac{2n}{\sqrt{3n^2 +2n}} = \frac{2}{\sqrt{3}} \)，所以所求為 \(\displaystyle \frac{2}{\sqrt{3}} \)
