每日一题[2119]两边夹

已知正整数数列 $\{a_n\}$ 满足：$a_4=16$，且

$$\dfrac1{a_{n+1}^2}<\dfrac1{a_n}+\dfrac1{a_{n+1}}-\dfrac2{n(n+1)}<\dfrac1{a_n^2},$$ 求数列 $\{a_n\}$ 的通项公式．

答案    $a_n=n^2$（$n\in\mathbb N^{\ast}$）

解析    根据题意，显然有 $\{a_n\}$ 单调递增，且有

$$\dfrac{1}{a_1^2}>\dfrac{1}{a_1}+\dfrac{1}{a_2}-1>\dfrac{1}{a_2^2}>\dfrac1{a_2}+\dfrac{1}{a_3}-\dfrac 13>\dfrac1{a_3^2}>\dfrac{1}{a_3}+\dfrac{1}{a_4}-\dfrac 16>\dfrac{1}{a_4^2},$$ 由 $a_4=16$，可得 $$\dfrac1{a_3^2}>\dfrac{1}{a_3}-\dfrac 5{48}>\dfrac{1}{256}\implies a_3=9,$$ 进而 $$\dfrac{1}{a_2^2}>\dfrac{1}{a_2}-\dfrac 29>\dfrac1{81}\implies a_2=4,$$ 进而 $$\dfrac{1}{a_1^2}>\dfrac{1}{a_1}-\dfrac 34>\dfrac1{16}\implies a_1=1.$$ 接下来利用数学归纳法证明，若 $a_k=k^2$（$k\geqslant 4$），那么 $a_{k+1}=(k+1)^2$．此时有 $$\dfrac{1}{a_{k+1}^2}<\dfrac{1}{k^2}+\dfrac{1}{a_{k+1}}-\dfrac{2}{k(k+1)}<\dfrac1{k^4},$$ 解得 $$\dfrac{k^4(k+1)}{k^3-k^2+k+1}<a_{k+1}<\dfrac{k^3+k^2+k\sqrt{k^4+2k^3-3k^2+4}}{2(k-1)},$$ 从而 $$-\dfrac{1+3k+2k^2}{1+k-k^2+k^3}<a_{k+1}-(k+1)^2<\dfrac{2(k+1)(2k+1)}{k\sqrt{k^4+2k^3-3k^2+4}+k^3+k^2-2k-2},$$ 由于 $$(k^3-k^2+k+1)-(2k^2+3k+1)=k^3-2k^2-2k>0,$$ 于是 $$-1<-\dfrac{1+3k+2k^2}{1+k-k^2+k^3}<0,$$ 且 $$0<\dfrac{2(k+1)(2k+1)}{k\sqrt{k^4+2k^3-3k^2+4}+k^3+k^2-2k-2}<\dfrac{2(k+1)(2k+1)}{2k^3}<1,$$ 因此 $a_{k+1}-(k+1)^2=0$，从而 $a_{k+1}=(k+1)^2$，命题得证． 综上所述，数列 $\{a_n\}$ 的通项公式为 $a_n=n^2$（$n\in\mathbb N^{\ast}$）．