$\lim_{n\to\infty}\sum_{k=1}^n f\left(\frac{k}{n^2}\right)=\frac{f'(0)}{2}. $
设 $f(x)$ 在 $(-1,1)$ 内有定义, 在 $x=0$ 处可导, 且 $f(0)=0$. 证明:
$\lim_{n\to\infty}\sum_{k=1}^n f\left(\frac{k}{n^2}\right)=\frac{f'(0)}{2}. $
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
$$\begin{aligned} \alpha(x)=\frac{f(x)-f(0)}{x}-f'(0), \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
则由导数定义知 $\displaystyle \lim_{x\to 0}\alpha(x)=0$, 且
$$\begin{aligned} f(x)=f(0)+f'(0)x+\alpha(x)x=f'(0)x+\alpha(x)x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
于是
$$\begin{aligned} f\left(\frac{k}{n^2}\right)&=f'(0)\frac{k}{n^2}+\alpha\left(\frac{k}{n^2}\right)\frac{k}{n^2},\\ \sum_{k=1}^n f\left(\frac{k}{n^2}\right)&=f'(0)\frac{n(n+1)}{2n^2} +\sum_{k=1}^n\alpha\left(\frac{k}{n^2}\right)\frac{k}{n^2},\\ \left|\sum_{k=1}^n f\left(\frac{k}{n^2}\right)-\frac{1}{2}f'(0)\right| &\leq \left|\frac{f'(0)}{2n}\right|+\sup_{x\in \left(0,\frac{1}{n}\right)} \alpha(x)\cdot \frac{n(n+1)}{2n^2}\\ &<\frac{|f'(0)|}{2n}+\frac{1}{2} \sup_{x\in \left(0,\frac{1}{n}\right)} \alpha(x)\\ &\to 0\left(n\to\infty\right). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{zhangzujin.cn}\end{array}}\end{aligned}$$
跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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