每日一题[785]求和放缩

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：$\dfrac{4}{a_1}+\dfrac{5}{a_2}+\cdots +\dfrac{n+3}{a_n}<\dfrac 43$．

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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．