问题及解答

求极限

\[
\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^3+1}+\frac{4}{n^3+2}+\cdots+\frac{n^2}{n^3+n}\biggr)
\]

注意到

\[
\frac{k^2}{n^3+n}\leqslant\frac{k^2}{n^3+k}\leqslant\frac{k^2}{n^3+1},
\]

因此

\[
\sum_{k=1}^{n}\frac{k^2}{n^3+n}\leqslant\sum_{k=1}^{n}\frac{k^2}{n^3+k}\leqslant\sum_{k=1}^{n}\frac{k^2}{n^3+1},
\]

即

\[
\frac{1}{n^3+n}\cdot\frac{1}{6}n(n+1)(2n+1)\leqslant\sum_{k=1}^{n}\frac{k^2}{n^3+k}\leqslant\frac{1}{n^3+1}\cdot\frac{1}{6}n(n+1)(2n+1)
\]

因此使用夹逼准则, 得

\[
\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{k^2}{n^3+k}=\frac{1}{3}.
\]