每日一题[724]数列求和

已知正项数列$\{a_n\}$满足$a_1=\dfrac 32$，$a_{n+1}^2-a_n^2=\dfrac{1}{(n+2)^2}-\dfrac{1}{n^2}$，记数列$\{a_n\}$的前$n$项和为$S_n$，则$\dfrac{1}{S_1}-\dfrac{1}{S_3}+\dfrac{1}{S_5}-\cdots -\dfrac{1}{S_{2007}}+\dfrac{1}{S_{2009}}$的值为______．

分析与解　$\dfrac{1006}{2011}$．

由累加法知 $$a_n^2=\dfrac 1{(n+1)^2}+\dfrac 1{n^2}+1,$$ 所以$a_n=1+\dfrac 1n-\dfrac{1}{n+1}$．直接求和得$S_n=n+1-\dfrac 1{n+1}$，所以 $$\dfrac 1{S_n}=\dfrac {n+1}{n(n+2)}=\dfrac 12\left(\dfrac 1n+\dfrac 1{n+2}\right).$$ 所求的式子记为$M$，有 $$2M=1+\dfrac 13-\left(\dfrac 13+\dfrac 15\right)+\left(\dfrac 15+\dfrac 17\right)-\left(\dfrac 17+\dfrac 19\right)+\cdots+\left(\dfrac 1{2009}+\dfrac 1{2011}\right)=\dfrac{2012}{2011},$$ 所以$M=\dfrac{1006}{2011}$．