已知 $m$ 为实数，数列 $\{a_n\}$ 的前 $n$ 项和为 $S_n$，满足：$S_n=\dfrac 98 a_n-\dfrac 43 \times 3^n+m$，且 $a_n \geqslant \dfrac{64}{3}$ 对任何的正整数 $n$ 恒成立．求证：当 $m$ 取到最大值时，对任何正整数 $n$ 都有 $\displaystyle \sum\limits_{k=1}^n{\dfrac{3^k}{S_k}}<\dfrac 3{16}$．
【难度】
【出处】
2011年全国高中数学联赛四川省预赛
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【答案】
略
【解析】
当 $n=1$ 时，由$$a_1=\dfrac 98 a_1-4+m$$得 $a_1=8(4-m)$．
当 $n\geqslant 1$ 时，因为\[\begin{split}&S_n=\dfrac 98 a_n-\dfrac 43 \times 3^n+m,\\&S_{n+1}=\dfrac 98a_{n+1}-\dfrac 43 \times 3^{n+1}+m,\end{split}\]所以$$a_{n+1}=\dfrac 98 a_{n+1}-\dfrac 98a_n-\dfrac 83 \times 3^n,$$即$$a_{n+1}=9a_n+\dfrac{64}{3}\times 3^n,$$所以$$a_{n+1}+\dfrac{32}{9}\times 3^{n+1}=9\left(a_n+\dfrac{32}{9}\times 3^n\right),$$故$$a_n+\dfrac{32}{9}\times 3^n=\left(a_1+\dfrac{32}{3}\right)\times 9^{n-1},$$即$$a_n=\dfrac 8{27}(16-3m)\times 9^n-\dfrac{32}{9}\times 3^n.$$由条件知，$$\dfrac 8{27}(16-3m)\times 9^n-\dfrac{32}{9}\times 3^n\geqslant \dfrac{64}{3}$$对任何正整数 $n$ 恒成立．
由于 $\dfrac{64}{3}\times \dfrac 1{9^n}+\dfrac{32}{9}\times \dfrac 1{3^n}$ 在 $n=1$ 时取得最大值为$$\dfrac{64}{3}\times \dfrac 19+\dfrac {32}{9}\times \dfrac 13 =\dfrac{96}{27}.$$于是$$\dfrac 8{27}(16-3m)\geqslant \dfrac{96}{27},$$解得 $m\leqslant \dfrac 43$．
由上式知道 $m$ 的最大值为 $\dfrac 43$．
当 $m=\dfrac 43$ 时，$$a_n=\dfrac{32}{9}\times 9^n-\dfrac{32}{9}\times 3^n,$$于是\[\begin{split}S_n&=\dfrac 98\left(\dfrac{32}{9}\times 9^n-\dfrac{32}{9}\times 3^n\right)-\dfrac 43 \times 3^n+\dfrac 43\\&=\dfrac 43[3\times (3^n)^2-4\times 3^n+1]\\&=\dfrac 43(3^{n+1}-1)(3^n-1),\end{split}\]所以\[\begin{split}\sum\limits_{k=1}^n{\dfrac{3^k}{S_k}}&=\dfrac 34 \sum\limits_{k=1}^n{\dfrac{3^k}{(3^{k+1}-1)(3^k-1)}}\\&=\dfrac 38\sum\limits_{k=1}^n\left(\dfrac 1{3^k-1}-\dfrac 1{3^{k+1}-1}\right)\\&=\dfrac 38\left(\dfrac 1{3-1}-\dfrac 1{3^{n+1}-1}\right)\\&<\dfrac 38\times \dfrac 12 =\dfrac 3{16}.\end{split}\]
当 $n\geqslant 1$ 时，因为\[\begin{split}&S_n=\dfrac 98 a_n-\dfrac 43 \times 3^n+m,\\&S_{n+1}=\dfrac 98a_{n+1}-\dfrac 43 \times 3^{n+1}+m,\end{split}\]所以$$a_{n+1}=\dfrac 98 a_{n+1}-\dfrac 98a_n-\dfrac 83 \times 3^n,$$即$$a_{n+1}=9a_n+\dfrac{64}{3}\times 3^n,$$所以$$a_{n+1}+\dfrac{32}{9}\times 3^{n+1}=9\left(a_n+\dfrac{32}{9}\times 3^n\right),$$故$$a_n+\dfrac{32}{9}\times 3^n=\left(a_1+\dfrac{32}{3}\right)\times 9^{n-1},$$即$$a_n=\dfrac 8{27}(16-3m)\times 9^n-\dfrac{32}{9}\times 3^n.$$由条件知，$$\dfrac 8{27}(16-3m)\times 9^n-\dfrac{32}{9}\times 3^n\geqslant \dfrac{64}{3}$$对任何正整数 $n$ 恒成立．
由于 $\dfrac{64}{3}\times \dfrac 1{9^n}+\dfrac{32}{9}\times \dfrac 1{3^n}$ 在 $n=1$ 时取得最大值为$$\dfrac{64}{3}\times \dfrac 19+\dfrac {32}{9}\times \dfrac 13 =\dfrac{96}{27}.$$于是$$\dfrac 8{27}(16-3m)\geqslant \dfrac{96}{27},$$解得 $m\leqslant \dfrac 43$．
由上式知道 $m$ 的最大值为 $\dfrac 43$．
当 $m=\dfrac 43$ 时，$$a_n=\dfrac{32}{9}\times 9^n-\dfrac{32}{9}\times 3^n,$$于是\[\begin{split}S_n&=\dfrac 98\left(\dfrac{32}{9}\times 9^n-\dfrac{32}{9}\times 3^n\right)-\dfrac 43 \times 3^n+\dfrac 43\\&=\dfrac 43[3\times (3^n)^2-4\times 3^n+1]\\&=\dfrac 43(3^{n+1}-1)(3^n-1),\end{split}\]所以\[\begin{split}\sum\limits_{k=1}^n{\dfrac{3^k}{S_k}}&=\dfrac 34 \sum\limits_{k=1}^n{\dfrac{3^k}{(3^{k+1}-1)(3^k-1)}}\\&=\dfrac 38\sum\limits_{k=1}^n\left(\dfrac 1{3^k-1}-\dfrac 1{3^{k+1}-1}\right)\\&=\dfrac 38\left(\dfrac 1{3-1}-\dfrac 1{3^{n+1}-1}\right)\\&<\dfrac 38\times \dfrac 12 =\dfrac 3{16}.\end{split}\]
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