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设数列 ${\displaystyle \{a_n\}}$ 的前 ${\displaystyle n}$ 项之积为 ${\displaystyle T_n}$, 满足 ${\displaystyle 2a_n+T_n=1(n\in {\mathbb{N}}^{\ast })}$.

1. 设 ${\displaystyle b_n=1+{\displaystyle \frac{1}{T_n}}}$, 求数列 ${\displaystyle \{b_n\}}$ 的通项公式 ${\displaystyle b_n}$;
2. 设数列 ${\displaystyle \{a_n\}}$ 的前 ${\displaystyle n}$ 项之和为 ${\displaystyle S_n}$, 证明: $${\displaystyle \frac{n}{2}-\frac{1}{2}+{\left(\frac{1}{2}\right)}^{n+1}<S_n<\frac{n}{2}+\frac{1}{2}\ln \frac{2^{n+1}}{2^{n+1}-1}-\frac{1}{4}.}$$

1. 解 ${\displaystyle \{b_n\}}$ 的表达式意味着 ${\displaystyle T_n\ne 0}$.当 ${\displaystyle n\ne 2}$,条件式意味着$$\begin{array}{r}2\frac{T_n}{T_{n-1}}+T_n=1\Rightarrow \frac{1}{T_n}+1=2(\frac{1}{T_{n-1}}+1),\end{array}$$
   再由 ${\displaystyle n=1}$ 解得 ${\displaystyle a_1=T_1={\displaystyle \frac{1}{3}}}$,从而 ${\displaystyle {\displaystyle \{\frac{1}{T_n}+1\}}}$ 是以 ${\displaystyle 4}$ 为首项，${\displaystyle 2}$ 为公比的等比数列，进而 ${\displaystyle {\displaystyle b_n=\frac{1}{T_n}+1=2^{n+1}}}$.
2. 证明 当 ${\displaystyle n\ge 2}$, ${\displaystyle a_n={\displaystyle \frac{T_n}{T_{n-1}}=\frac{2^n-1}{2^{n+1}-1}}}$ (这事实上也是 ${\displaystyle n=1}$ 的情形).
   一方面$$\begin{array}{r}{a}_n=\frac{1}{2}-\frac{1}{2^{n+2}-2}>\frac{1}{2}-\frac{1}{2^{n+2}-2^{n+1}}=\frac{1}{2}-\frac{1}{2^{n+1}},\end{array}$$
   累加即得$$\begin{array}{r}{S}_n>\sum _{i=1}^n(\frac{1}{2}-\frac{1}{2^{n+1}})=\frac{n}{2}-\frac{1}{2}+{\left(\frac{1}{2}\right)}^{n+1}.\end{array}$$
   另一方面$$\begin{array}{r}{a}_n=\frac{1}{2}-\frac{1}{2^{n+2}}<\frac{1}{2}-\frac{1}{2^{n+2}},\end{array}$$
   故累加得$$\begin{array}{r}{S}_n<\frac{n}{2}-\frac{1}{4}+\frac{1}{2^{n+2}}.\end{array}$$
   又$$\begin{array}{rl}& \ln \frac{2^{n+1}}{2^{n+1}-1}>1-\frac{2^{n+1}-1}{2^{n+1}}=\frac{1}{2^{n+1}}\\ \Rightarrow & S_n<\frac{n}{2}-\frac{1}{4}+\frac{1}{2}\ln \frac{2^{n+1}}{2^{n+1}-1}.\end{array}$$
   因此不等式得证.