设$a_n=n(n+1)\cdot 2^n$，$n\in\mathbb N^*$．
(1) 求证：$\dfrac{3}{a_1}+\dfrac{4}{a_2}+\cdots +\dfrac{n+2}{a_n}<1$；
(2) 求证：$\dfrac{4}{a_1}+\dfrac{5}{a_2}+\cdots +\dfrac{n+3}{a_n}<\dfrac 43$．
证明 (1) 由于$$\dfrac{n+2}{a_n}=\dfrac{n+2}{n(n+1)\cdot 2^n}=\dfrac{1}{n\cdot 2^{n-1}}-\dfrac{1}{(n+1)\cdot 2^n},$$于是$$\dfrac{3}{a_1}+\dfrac{4}{a_2}+\cdots +\dfrac{n+2}{a_n}=1-\dfrac{1}{(n+1)\cdot 2^n}<1,$$原命题得证．
(2)裂项放缩 由于$$\dfrac{n+3}{n(n+1)\cdot 2^n}<\dfrac{1}{\left(n-\dfrac 14\right)\cdot 2^{n-1}}-\dfrac{1}{\left(n+\dfrac 34\right)\cdot 2^n},$$于是$$\dfrac{4}{a_1}+\dfrac{5}{a_2}+\cdots +\dfrac{n+3}{a_n}<\dfrac 43-\dfrac{1}{\left(n+\dfrac 34\right)\cdot 2^n}<\dfrac 43,$$原命题得证．
等比放缩 注意到当$n\geqslant 3$时，有$$\dfrac{n+3}{n(n+1)}\leqslant \dfrac 12,$$于是有
\[\begin{split} LHS&<1+\dfrac 5{24}+\dfrac 12\cdot \dfrac{1}{2^3}+\cdots +\dfrac 12\cdot \dfrac{1}{2^n}+\cdots \\&=\dfrac{29}{24}+\dfrac{\dfrac{1}{16}}{1-\dfrac 12}\\&=\dfrac 43,\end{split} \]原命题得证．