每日一题[2464]回归定义

$\displaystyle\lim_{n\to +\infty}\sum_{k=1}^n\dfrac{n}{(2n-k)(2n+k)}=$_______．

答案    $\dfrac14\ln3$．

解析    根据题意, 有

$$\begin{split} \lim_{n\to +\infty}\sum_{k=1}^n\dfrac{n}{(2n-k)(2n+k)}&=\lim_{n\to +\infty}\left(\dfrac 1n\cdot \sum_{k=1}^n\dfrac{1}{\left(2-\dfrac kn\right)\left(2+\dfrac kn\right)}\right)\\ &=\int_0^1\dfrac{1}{(2-x)(2+x)}{ {\rm d}} x\\ &=\dfrac14\int_0^1\left(\dfrac{1}{2-x}+\dfrac{1}{2+x}\right){ {\rm d}} x\\ &=\dfrac14\left(\ln |2+x|-\ln |2-x|\right)\Big|_0^1\\ &=\dfrac 14\ln 3.\end{split}$$