計算2，
\(a_n=(1^2+2^2+\ldots+n^2)(1^5+2^5+\ldots+n^5)\)，
\(b_n=(1^3+2^3+\ldots+n^3)(1^4+2^4+\ldots+n^4)\)
則\(\displaystyle \lim_{n\to \infty}\frac{a_n}{b_n}=\)為何？
[解答]
同除 \( n^9 \)，再用黎曼和
\(\displaystyle \frac{a_{n}}{n^{9}}=\frac{1^{2}+2^{2}+\ldots+n^{2}}{n^{3}}\times\frac{1^{5}+2^{5}+\ldots+n^{5}}{n^{6}}=\frac{1}{n}\left((\frac{1}{n})^{2}+(\frac{2}{n})^{2}+\ldots+(\frac{n}{n})^{2}\right)\times\frac{1}{n}\left((\frac{1}{n})^{5}+(\frac{2}{n})^{5}+\ldots+(\frac{n}{n})^{5}\right) \)
故 \(\displaystyle \lim\limits _{n\to\infty}\frac{a_{n}}{n^{9}}=\int_{0}^{1}x^{2}dx\int_{0}^{1}x^{5}dx=\frac{1}{3\times6}=\frac{1}{18} \)
同理 \(\displaystyle \lim\limits _{n\to\infty}\frac{b_{n}}{n^{9}}=\int_{0}^{1}x^{3}dx\int_{0}^{1}x^{4}dx=\frac{1}{4\times5}=\frac{1}{20} \)
故所求 \( = \frac{1}{18} \div \frac{1}{20} = \frac{10}{9} \)
111.2.14補充
\(\displaystyle \lim_{n\to \infty}\frac{(1^2+2^2+3^2+\ldots+n^2)(1^5+2^5+3^5+\ldots+n^5)}{(1^3+2^3+3^3+\ldots+n^3)(1^4+2^4+3^4+\ldots+n^4)}=\)？
(105萬芳高中，
https://math.pro/db/viewthread.php?tid=2542&page=1#pid16610)
