问题及解答

利用定积分求极限

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

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

\[
U_n:=\sum_{k=1}^{n}\frac{1}{1+(\frac{k+1}{n})^2}\cdot\frac{1}{n} < S_n < \sum_{k=1}^{n}\frac{1}{1+(\frac{k}{n})^2}\cdot\frac{1}{n}=:V_n 
\]

其中

\[
\lim_{n\rightarrow\infty}V_n=\int_0^1\frac{1}{1+x^2}\mathrm{d}x=\arctan x\biggr|_{0}^{1}=\frac{\pi}{4}.
\]

而

\[
\begin{split}
U_n&=\sum_{i=2}^{n+1}\frac{1}{1+(\frac{i}{n})^2}\cdot\frac{1}{n}=\sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^2}\cdot\frac{1}{n}+\biggl(\frac{1}{1+(\frac{n+1}{n})^2}\cdot\frac{1}{n}-\frac{1}{1+(\frac{1}{n})^2}\cdot\frac{1}{n}\biggr)\\
&=V_n+\biggl(\frac{1}{n+\frac{(n+1)^2}{n}}-\frac{1}{n+\frac{1}{n}}\biggr),
\end{split}
\]

于是

\[
\lim_{n\rightarrow\infty}U_n=\lim_{n\rightarrow\infty}V_n+0=\frac{\pi}{4}.
\]

因此

\[
\lim_{n\rightarrow\infty}S_n=\frac{\pi}{4}.
\]