求证：$\displaystyle\sum_{k=1}^n\dfrac{k+1}{k^2(k+2)^2}<\dfrac{5}{16}$．
解析
根据题意，有\[\sum_{k=1}^n\dfrac{k+1}{k^2(k+2)^2}=\sum_{k=1}^n\left(\dfrac{1}{4k^2}-\dfrac{1}{4(k+2)^2}\right)=\dfrac 14+\dfrac{1}{16}-\dfrac{1}{4(n+1)^2}-\dfrac{1}{4(n+2)^2}<\dfrac 5{16}.\]