数学分析考研每日一题002:迫敛性与积分法

(南京师范大学,2022)求极限 {\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)^{\frac{1}{n}}}.

解答:对任意的正整数 {n},显然有 {1 \leq 1+\frac{1}{2}+\cdots+\frac{1}{n} \leq n},于是
{1 \leq\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)^{\frac{1}{n}} \leq n^{\frac{1}{n}} . }
而 {\lim _{n \rightarrow \infty} n^{\frac{1}{n}}=1},所以由迫敛性可知 {\lim _{n \rightarrow \infty}\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)^{\frac{1}{n}}=1}.

(北京工业大学,2022)求极限{I=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}+k}.}

解答:对任意的正整数 {n},当 {1 \leq k \leq n} 时,明显有 {\frac{n}{(n+1)^{2}+k^{2}} \leq \frac{n}{n^{2}+k^{2}+k} \leq \frac{n}{n^{2}+k^{2}}},进而
\begin{equation} \sum_{k=1}^{n} \frac{n}{(n+1)^{2}+k^{2}} \leq \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}+k} \leq \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}} . \label{polianxing}(*) \end{equation}
而由定积分的性质,明显有
{\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+\left(\frac{k}{n}\right)^{2}}=\int_{0}^{1} \frac{1}{1+x^{2}} \mathrm{d} x=\left.\arctan x\right|_{0}^{1}=\frac{\pi}{4} . }
再结合 {\lim _{n \rightarrow \infty} \frac{n}{(n+1)^{2}+(n+1)^{2}}=0} 还有
{\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{n}{(n+1)^{2}+k^{2}}=\lim _{n \rightarrow \infty} \sum_{k=1}^{n+1} \frac{n}{(n+1)^{2}+k^{2}}=\lim _{n \rightarrow \infty} \frac{n}{n+1} \lim _{n \rightarrow \infty} \sum_{k=1}^{n+1} \frac{n+1}{(n+1)^{2}+k^{2}}=1 \cdot \frac{\pi}{4}=\frac{\pi}{4} . }
所以根据(*)式,由迫敛性可知
{I=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{n}{n^{2}+k^{2}+k}=\frac{\pi}{4} . }

(西南大学,2022)求极限 {\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{2 n^{2}+1}}+\frac{1}{\sqrt{2 n^{2}+2}}+\cdots+\frac{1}{\sqrt{2 n^{2}+n}}\right)}.

解答:由于对任意的正整数 {n},当 {1 \leq k \leq n} 时,有
{\frac{1}{\sqrt{2 n^{2}+n}} \leq \frac{1}{\sqrt{2 n^{2}+k}} \leq \frac{1}{\sqrt{2 n^{2}+1}} . }
所以
{\frac{n}{\sqrt{2 n^{2}+n}} \leq \frac{1}{\sqrt{2 n^{2}+1}}+\frac{1}{\sqrt{2 n^{2}+2}}+\cdots+\frac{1}{\sqrt{2 n^{2}+n}} \leq \frac{n}{\sqrt{2 n^{2}+1}} . }
而明显

\lim _{n \rightarrow \infty} \frac{n}{\sqrt{2 n^{2}+n}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{2+\frac{1}{n}}}=\frac{1}{\sqrt{2}}
\lim _{n \rightarrow \infty} \frac{n}{\sqrt{2 n^{2}+1}}=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{2+\frac{1}{n^{2}}}}=\frac{1}{\sqrt{2}}

所以由迫敛性可得
{\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{2 n^{2}+1}}+\frac{1}{\sqrt{2 n^{2}+2}}+\cdots+\frac{1}{\sqrt{2 n^{2}+n}}\right)=\frac{1}{\sqrt{2}} }

(兰州大学,2022)求极限 {\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}+k^{4}}}.

解答:首先对任意的正整数 {n},当 {1 \leq k \leq n} 时,有
{\frac{(2 k-1)^{4}}{n^{5}+n^{4}} \leq \frac{(2 k-1)^{4}}{n^{5}+k^{4}} \leq \frac{(2 k-1)^{4}}{n^{5}} . }
上式关于 {k} 求和,就有
\frac{n^{5}}{n^{5}+n^{4}} \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}} \leq \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}+k^{4}} \leq \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}} . \label{polianxing2}(**)
而由定积分的性质可知
{\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}}=\frac{1}{2} \lim _{n \rightarrow \infty} \frac{2}{n} \sum_{k=1}^{n}\left(\frac{2 k-1}{n}\right)^{4}=\frac{1}{2} \int_{0}^{2} x^{4} \mathrm{d} x=\frac{16}{5} . }
再结合 {\lim _{n \rightarrow \infty} \frac{n^{5}}{n^{5}+n^{4}}=\lim _{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}=1} 可知
{\lim _{n \rightarrow \infty} \frac{n^{5}}{n^{5}+n^{4}} \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}}=\lim _{n \rightarrow \infty} \frac{n^{5}}{n^{5}+n^{4}} \cdot \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}}=\frac{16}{5} . }
所以结合(**)式,由迫敛性可知 {\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(2 k-1)^{4}}{n^{5}+k^{4}}=\frac{16}{5}}.

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