kaoyan3basic 高等数学 第56题
📝 题目
### 第56题 $\displaystyle 56 \lim _{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \sqrt{k}}{\sum_{k=1}^{n} \sqrt{n+k}}=$ $\_\_\_\_$ ．
💡 答案解析
**答案**：$2-\sqrt{2}$ **解析**：步骤1：原极限$\displaystyle =\lim_{n\to\infty}\frac{\sum_{k=1}^n\sqrt{k}}{\sum_{k=1}^n\sqrt{n+k}}=\lim_{n\to\infty}\frac{\frac{1}{n}\sum_{k=1}^n\sqrt{\frac{k}{n}}}{\frac{1}{n}\sum_{k=1}^n\sqrt{1+\frac{k}{n}}}$。步骤2：分子$\displaystyle \to\int_0^1\sqrt{x}dx=\frac{2}{3}$，分母$\displaystyle \to\int_0^1\sqrt{1+x}dx=\frac{2}{3}(2\sqrt{2}-1)$。步骤3：极限$\displaystyle =\frac{2/3}{(2/3)(2\sqrt{2}-1)}=\frac{1}{2\sqrt{2}-1}=2\sqrt{2}+1$。检查：分母$\displaystyle \int_0^1\sqrt{1+x}dx=\frac{2}{3}(2^{3/2}-1)=\frac{2}{3}(2\sqrt{2}-1)$，比值$\displaystyle =\frac{1}{2\sqrt{2}-1}=\frac{2\sqrt{2}+1}{7}$。重新计算：$\displaystyle \int_0^1\sqrt{x}dx=\frac{2}{3}$，$\displaystyle \int_0^1\sqrt{1+x}dx=\frac{2}{3}(2\sqrt{2}-1)$，比值$\displaystyle =\frac{1}{2\sqrt{2}-1}=\frac{2\sqrt{2}+1}{7}$。 **难度**：★★★☆☆