已知数列 $\left\{ {a_n}\right\} $ 满足 ${a_1} = 1$，点 $\left({a_n} , {a_{n + 1}}\right)$ 在直线 $y = 2x + 1$ 上．
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求数列 $\left\{ {a_n}\right\} $ 的通项公式；标注答案${a_n} = {2^n} - 1\left(n \in {{\mathbb{N}}^ * }\right)$解析因为点 $\left({a_n} , {a_{n + 1}}\right)$ 在直线 $y = 2x + 1$ 上，所以\[{a_{n + 1}} = 2{a_n} + 1,\]上式两端加 $ 1 $，得\[{a_{n + 1}} + 1 = 2\left({a_n} + 1\right),\]则数列 $\left\{ {a_n} + 1\right\} $ 是以 $2$ 为首项，$2$ 为公比的等比数列，所以$${a_n} = {2^n} - 1\left(n \in {{\mathbb{N}}^ * }\right).$$
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若数列 $\left\{ {b_n}\right\} $ 满足 ${b_1} = {a_1} $，$ \dfrac{b_n}{a_n} = \dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{{{a_{n - 1}}}}\left(n \geqslant 2 , n \in {{\mathbb{N}}^*}\right)$．
求 ${b_{n + 1}}{a_n} - \left({b_n} + 1\right){a_{n + 1}}$ 的值；标注答案$n=1$ 时，${b_{n + 1}}{a_n} - \left({b_n} + 1\right){a_{n + 1}} = -3$；
$n\geqslant 2$ 时，${b_{n + 1}}{a_n} - \left({b_n} + 1\right){a_{n + 1}} = 0 $解析因为\[\dfrac{b_n}{a_n} = \dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{{{a_{n - 1}}}},\]所以\[\dfrac{{{b_{n + 1}}}}{{{a_{n + 1}}}}= \dfrac{b_{n}}{a_{n}}+\dfrac{1}{a_{n}},\]于是当 $ n \geqslant 2 $ 且 $ n \in {{\mathbb{N}}^ * } $ 时，\[{b_{n + 1}}{a_n} - \left({b_n} + 1\right){a_{n + 1}} = 0 .\]当 $n = 1$ 时，\[{b_2}{a_1} - \left({b_1} + 1\right){a_2} = -a_2a_1=- 3.\] -
对于 $(2)$ 中的数列 $\left\{ {b_n}\right\} $，求证：$\left(1 + {b_1}\right)\left(1 + {b_2}\right) \cdots \left(1 + {b_n}\right) < \dfrac{10}{3}{b_1}{b_2} \cdots {b_n}$ $\left(n \in {{\mathbb{N}}^*}\right)$．标注答案略解析由 $(2)$ 知\[\dfrac{{{b_n} + 1}}{{{b_{n + 1}}}} = \dfrac{a_n}{{{a_{n + 1}}}}\left(n \geqslant 2\right) , {b_2} = {a_2}.\]所以\[\begin{split} &\left( {1 + \dfrac{1}{b_1}} \right)\left( {1 + \dfrac{1}{b_2}} \right) \cdots \left( {1 + \dfrac{1}{b_n}} \right)\\=& \dfrac{{{b_1} + 1}}{b_1} \cdot \dfrac{{{b_2} + 1}}{b_2} \cdots \dfrac{{{b_n} + 1}}{b_n} \\= &\dfrac{1}{b_1} \cdot \dfrac{{{b_1} + 1}}{b_2} \cdot \dfrac{{{b_2} + 1}}{b_3} \cdots \dfrac{{{b_{n - 1}} + 1}}{b_n} \cdot \dfrac{{{b_n} + 1}}{{{b_{n + 1}}}} \cdot {b_{n + 1}}\\= &\dfrac{1}{b_1} \cdot \dfrac{{{b_1} + 1}}{b_2} \cdot \dfrac{a_2}{a_3} \cdot \dfrac{a_3}{a_4} \cdots \dfrac{{{a_{n - 1}}}}{a_n} \cdot \dfrac{a_n}{{{a_{n + 1}}}} \cdot {b_{n + 1}} \\ =& 2 \cdot \dfrac{{{b_{n + 1}}}}{{{a_{n + 1}}}} \\=& 2\left(\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_n}\right).\end{split}\]因为 $k \geqslant 2$ 时，\[\begin{split}\dfrac{1}{{{2^k} - 1}} &= \dfrac{{{2^{k + 1}} - 1}}{{\left({2^k} - 1\right)\left({2^{k + 1}} - 1\right)}}\\ <& \dfrac{{{2^{k + 1}}}}{{\left({2^k} - 1\right)\left({2^{k + 1}} - 1\right)}} \\=& 2\left(\dfrac{1}{{{2^k} - 1}} - \dfrac{1}{{{2^{k + 1}} - 1}}\right),\end{split} \]所以\[\begin{split} &\dfrac{1}{a_1} + \dfrac{1}{a_2} + \cdots + \dfrac{1}{a_n} \\= &1 + \dfrac{1}{3} + \cdots + \dfrac{1}{{{2^n} - 1}}\\
< &1 + 2\left[ {\left( {\dfrac{1}{{{2^2} - 1}} - \dfrac{1}{{{2^3} - 1}}} \right) + \cdots + \left( {\dfrac{1}{{{2^n} - 1}} - \dfrac{1}{{{2^{n + 1}} - 1}}} \right)} \right] \\=& 1 + 2\left( {\dfrac{1}{3} - \dfrac{1}{{{2^{n + 1}} - 1}}} \right) < \dfrac{5}{3} ,\end{split} \]于是\[\left( {1 + \dfrac{1}{b_1}} \right)\left( {1 + \dfrac{1}{b_2}} \right) \cdots \left( {1 + \dfrac{1}{b_n}} \right) < \dfrac{10}{3},\]即$$\left(1 + {b_1}\right)\left(1 + {b_2}\right) \cdots \left(1 + {b_n}\right) < \dfrac{10}{3}{b_1}{b_2} \cdots {b_n}.$$
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