题目

lim _(narrow infty )(dfrac (1)(sqrt {{n)^2+(1)^2}}+dfrac (1)(sqrt {{n)^2+(2)^2}}+... +dfrac (1)(sqrt {{n)^2+(n)^2}})= )=______.

$\lim\limits_{n\to\infty}\Bigg(\frac{1}{\sqrt{n^2+1^2}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots+\frac{1}{\sqrt{n^2+n^2}}\Bigg)=$

______.

题目解答

答案

根据所证明的极限表达式，利用定积分的定义，进行计算可以得到

$\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2^2}}+\cdots+\frac{1}{\sqrt{n^2+n^2}}\right)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\left(\frac{k}{n}\right)^2}}$

$\begin{aligned} =\int_0^1\frac{\mathrm{d}x}{\sqrt{1+x^2}}=\ln(1+\sqrt{2}). \end{aligned}$