已知数列 \(\{a_n\}\) 满足, \(a_1=1,a_{n+1}=2a_n+1\), \(n\in\mathbb{N}^\ast\).
\((1)\) 求 \(\{a_n\}\) 的通项公式.
\((2)\) \(\{ b_n\}\) 满足 \(4^{b_1-1}\cdot 4^{b_2-1} \cdots 4^{b_n-1}=\left(a_n+1 \right)^{b_n}\), 求证: \(\{ b_n\}\) 是等差数列.
\((3)\) 证明\(:\) $ \forall n\in\mathbb{N}^\ast,\dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots + \dfrac{1}{a_{n+1}}<\dfrac 23$.
解析:
\((1)\) 由题有\[ \forall n\in \mathbb{N}^\ast, a_{n+1}+1=2\left( a_n+1\right). \]
因此 \[ a_n=(a_1+1)\cdot 2^{n-1}-1=2^n-1,n\in\mathbb{N}^\ast.\]
\((2)\) 由题易知\[ 4^{b_1+b_2+\cdots +b_n-n}=2^{nb_n}=4^{\frac{1}{2}n b_n}.\]
因此 \(\displaystyle \sum_{k=1}^{n}b_k=\dfrac12nb_n+n\). 进而 \[ \forall n\in\mathbb{N}^\ast, b_{n+1}=\sum_{k=1}^{n+1}b_{k}-\sum_{k=1}^{n}b_k=\dfrac12(n+1)b_{n+1}-\dfrac 12nb_n+1.\]
整理得 \(nb_n=(n-1)b_{n+1}+2,n\in\mathbb{N}^\ast\), 从而
\[ \forall n\geqslant 2, \dfrac{b_{n+1}-2}{n}=\dfrac{b_n-2}{n-1}=\cdots =\dfrac{b_2-2}{2-1} .\]
所以 \(\forall n\geqslant 2, b_n=(b_2-2)n+2\). 又 \[ b_2-b_1=b_2-2,\]
因此数列 \(\{ b_n\}\) 是以 \(b_1=2\) 为首项, 以 \(b_2-2\) 为公差的等差数列.
\((3)\) 由于 \[ \dfrac{a_{n+1}}{a_n}=\dfrac{2^{n+1}-1}{2^n-1}=2+\dfrac{1}{2^n-1}>2.\]
所以 \(\dfrac{1}{a_{n+1}}<\dfrac 12\cdot \dfrac{ 1}{a_n}\), 记待证不等式左侧为 \(T_n\),
情形一 若 \(n=1\), 此时\[ T_1=\dfrac{1}{a_2}=\dfrac13<\dfrac23.\]
情形二 若 \(n\geqslant 2\), 此时\[ T_n\leqslant \dfrac{1}{a_2}+\dfrac{1}{a_3}\cdot \left( 1+\dfrac12+\cdots +\dfrac{1}{2^{n-2}}\right)<\dfrac{1}{3}+\dfrac27=\dfrac{13}{21}<\dfrac23.\]
综上, \(\forall n\in\mathbb{N}^\ast, T_n<\dfrac 23\). 证毕.
