Site Overlay

高等数学:无限求和极限题型

通过迫敛性和数列求和解决

  1. 求: $\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n}{n^{2}+n+n}\right)$

$\frac{i}{n^{2}+n+n} \leqslant \frac{i}{n^{2}+n+i} \leqslant \frac{i}{n^{2}+n+1}(i=1,2, \cdots, n)$

$\frac{\frac{1}{2} n(n+1)}{n^{2}+n+n} \leqslant \frac{1}{n^{2}+n+1}+\frac{2}{n^{2}+n+2}+\cdots+\frac{n}{n^{2}+n+n} \leqslant \frac{\frac{1}{2} n(n+1)}{n^{2}+n+1}$

$\lim {x \rightarrow \infty} \frac{\frac{1}{2} n(n+1)}{n^{2}+n+n}=\lim {n \rightarrow \infty} \frac{\frac{1}{2} n(n+1)}{n^{2}+n+1}=\frac{1}{2}$

所以由迫敛定理, 原式$=\frac{1}{2}$

  1. 求 $\lim _{x \rightarrow \infty}\left(\frac{1}{\sqrt{4 n^{2}+1}}+\frac{1}{\sqrt{4 n^{2}+2}}+\cdots+\frac{1}{\sqrt{4 n^{2}+n}}\right)$

因为$\frac{1}{\sqrt{4 n^{2}+n}} \leqslant \frac{1}{\sqrt{4 n^{2}+i}} \leqslant \frac{1}{\sqrt{4 n^{2}+1}}(i=1,2, \cdots, n)$

所以$\frac{n}{\sqrt{4 n^{2}+n}} \leqslant \frac{1}{\sqrt{4 n^{2}+1}}+\frac{1}{\sqrt{4 n^{2}+2}}+\cdots+\frac{1}{\sqrt{4 n^{2}+n}} \leqslant \frac{n}{\sqrt{4 n^{2}+1}}$

又因为$\lim {n \rightarrow \infty} \frac{n}{\sqrt{4 n^{2}+n}}=\lim {n \rightarrow \infty} \frac{n}{\sqrt{4 n^{2}+1}}=\frac{1}{2}$

所以由迫敛定理, 原式$=\frac{1}{2}$

通过积分解决

  1. $\lim _{n \rightarrow \infty} \frac{1^{2}+2^{2}+\cdots+n^{2}}{n^{3}}$

    提出一个 $\frac{1}{n}$ 然后转化为积分即可

    $\lim {n \rightarrow \infty} \frac{1^{2}+2^{2}+\cdots+n^{2}}{n^{3}}=\lim {n \rightarrow \infty} \frac{1}{n} \sum{i=1}^{n}\left(\frac{i}{n}\right)^{2}=\int{0}^{1} x^{2} \mathrm{d} x=\frac{1}{3}$

  2. $\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}+1^{2}}}+\frac{1}{\sqrt{n^{2}+2^{2}}}+\cdots+\frac{1}{\sqrt{n^{2}+n^{2}}}\right)$

    提出一个 $\frac{1}{n}$ 然后转化为积分即可

    $\lim {n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}+1^{2}}}+\frac{1}{\sqrt{n^{2}+2^{2}}}+\cdots+\frac{1}{\sqrt{n^{2}+n^{2}}}\right)=\lim {n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \frac{1}{\sqrt{1+\left(\frac{i}{n}\right)^{2}}}$

    $\lim {n \rightarrow \infty}\left(\frac{1}{\sqrt{n^{2}+1^{2}}}+\frac{1}{\sqrt{n^{2}+2^{2}}}+\cdots+\frac{1}{\sqrt{n^{2}+n^{2}}}\right)=\lim {n \rightarrow \infty} \frac{1}{n} \sum{i=1}^{n} \frac{1}{\sqrt{1+\left(\frac{i}{n}\right)^{2}}}$
    $=\int
    {0}^{1} \frac{\mathrm{d} x}{\sqrt{1+x^{2}}}=\left.\ln (x+\sqrt{1+x^{2}})\right|_{0} ^{1}=\ln (1+\sqrt{2})$

  3. $\lim _{n \rightarrow \infty}\left(\frac{\sin \frac{\pi}{n}}{n+1}+\frac{\sin \frac{2 \pi}{n}}{n+\frac{1}{2}}+\cdots+\frac{\sin \frac{n \pi}{n}}{n+\frac{1}{n}}\right)$

    先用迫敛性, 然后积分即可. 不过这里左边的被迫敛的那个玩意儿不太好找. 我姑且用这样的规律: 分母变小, 左对左. 分母变小, 左对右.

    $\frac{1}{n+1} \sum{i=1}^{n} \sin \frac{\pi i}{n} \leqslant \frac{\sin \frac{\pi}{n}}{n+1}+\frac{\sin \frac{2 \pi}{n}}{n+\frac{1}{2}}+\cdots+\frac{\sin \frac{n \pi}{n}}{n+\frac{1}{n}} \leqslant \frac{1}{n} \sum{i=1}^{n} \sin \frac{\pi i}{n}$

    $\begin{aligned} \lim {n \rightarrow \infty} \frac{1}{n+1} \sum{i=1}^{n} \sin \frac{\pi i}{n} &=\lim {n \rightarrow \infty} \frac{n}{n+1} \cdot \frac{1}{n} \sum{i=1}^{n} \sin \frac{\pi i}{n}=\lim {n \rightarrow \infty} \frac{1}{n} \sum{i=1}^{n} \sin \frac{\pi i}{n}=\int{0}^{1} \sin \pi x \mathrm{d} x \ &=\frac{1}{\pi} \int{0}^{\pi} \sin x \mathrm{d} x=\frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \sin x \mathrm{d} x=\frac{2}{\pi} \end{aligned}$

    $\lim _{n \rightarrow \infty}\left(\frac{\sin \frac{\pi}{n}}{n+1}+\frac{\sin \frac{2 \pi}{n}}{n+\frac{1}{2}}+\cdots+\frac{\sin \frac{n \pi}{n}}{n+\frac{1}{n}}\right)=\frac{2}{\pi}$

  4. $\dfrac{ \sqrt[n]{n !}}{n}$

    $\lim {n \rightarrow \infty} \frac{\sqrt[n]{n !}}{n}=\lim {n \rightarrow \infty}\left(\frac{1}{n} \cdot \frac{2}{n} \cdots \cdot \frac{n}{n}\right)^{\frac{1}{n}}=e^{\lim {n \rightarrow \infty} \frac{1}{n} \sum{i=1}^{n} \ln \frac{i}{n}}=e^{\int_{0}^{1} \ln x d x}=e^{-1}$

    此处要掌握处理阶乘的技巧: 拆! 另外还有: $n = (n^n)^{\frac{1}{n}} = (n\cdot n\cdot n\cdot \dots n\cdot)^{\frac{1}{n}}$

  5. $\lim {n \rightarrow \infty} \sum{i=1}^{n} \frac{1}{n+\frac{i^{2}+1}{n}}$
    $\sum{i=1}^{n} \frac{1}{n+\frac{(i+1)^{2}}{n}} \leqslant \sum{i=1}^{n} \frac{1}{n+\frac{i^{2}+1}{n}} \leqslant \sum{i=1}^{n} \frac{1}{n+\frac{i^{2}}{n}}=\frac{1}{n} \sum{i=1}^{n} \frac{1}{1+\left(\frac{i}{n}\right)^{2}}$

    $\sum{=1}^{n} \frac{1}{n+\frac{(i+1)^{2}}{n}}=\sum{i=1}^{n} \frac{1}{n+\frac{i^{2}}{n}}-\frac{1}{n+\frac{1}{n}}+\frac{1}{n+\frac{(n+1)^{2}}{n}}$

    $\lim {n \rightarrow \infty} \frac{1}{n} \sum{i=1}^{n} \frac{1}{1+\left(\frac{i}{n}\right)^{2}}=\int_{0}^{1} \frac{\mathrm{d} x}{1+x^{2}}=\frac{\pi}{4}$

发表评论

电子邮件地址不会被公开。 必填项已用*标注