«极限+求和=>积分»
by pluvet on Dec 22, 2019

极限+求和=>积分

例: 求 $\lim _{n \to\infty} \frac{\sum_{k=1}^{n} \sqrt{n^{2}-k^{2}}}{n^{2}}$

利用性质

$$ \int_0^a f(x) dx \approx \sum_{k=0}^{na} \frac{1}{n}f\left(\frac{k}{n}\right)\\ $$

原极限化为

$$ \frac{1}{n} \cdot\left(\sum_{k=1}^{n} \sqrt{1-\left(\frac{k}{n}\right)^{2}}\right)=\int_{0}^{1} \sqrt{1-x^{2}} $$

$y^{2}=1-x^{2}$
$\Rightarrow x^{2}+y^{2}=1$

image-20191222175517036.png

例: 求 $\lim\limits_{n \to \infty}\sum\limits_{k=0}^n \dfrac{\sqrt{n}}{n+k^2}(n=1,2,\cdots)$

$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} \frac{\sqrt{n}}{1+\frac{k}{\sin } j^{2}}$

let $t=\sqrt{n}$

$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n} \frac{t}{1+\left(\frac{k}{4}\right)^{2}}$
$=\lim _{n \rightarrow \infty} \frac{1}{t} \sum_{k=0}^{n} \frac{1}{1+\left(\frac{k}{4}\right)^{2}}$
$=\int_{0}^{1} \frac{dx^2}{1+x^{2}}=2 \arctan (1)=\frac{\pi}{4} \times 2=\frac{\pi}{2}$

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