引用:
原帖由 meifang 於 2012-6-26 12:06 AM 發表 
我想問一下第16的類題
\(\displaystyle \lim_{n \to \infty }\frac{1}{n^{2}}\sum_{k=1}^{n}\sqrt{k(k+2)}=\frac{1}{2}\) (97中和高中)
\(\displaystyle \lim_{n \to \infty }\frac{1}{\sqrt{n+1}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=2\) (97台南女中)
(1)\(\displaystyle \lim_{n \to \infty }\frac{1}{n^{2}}\sum_{k=1}^{n}\sqrt{k(k+2)}=\lim_{n \to \infty }\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{k}{n}(\frac{k}{n}+\frac{2}{n})}\)
\(=\int^{1}_{0}\sqrt{x^2}dx=\frac{1}{2}\)
(2)\(\displaystyle \lim_{n \to \infty }\frac{1}{\sqrt{n+1}}\sum_{k=1}^{n}\frac{1}{\sqrt{k}}=\lim_{n \to \infty }\frac{1}{n}\sum_{k=1}^{n}\frac{n}{\sqrt{k(n+1)}}\)
\( \displaystyle =\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\sqrt{\frac{n}{k}\times \frac{n}{n+1}}=\int^{1}_{0}\sqrt{\frac{1}{x}}dx=2\)
不過，基本上你應該去爬一下文
