已知数列 $\left\{ {a_n} \right\}$，$a_n \geqslant 0$，${a_1}=0$，$a_{n+1}^2+a_{n+1}-1=a_n^2$．记\[\begin{split} {S_n}&=a_1+a_2+\cdots +a_n,\\ {T_n}&=\dfrac{1}{{1+{a_1}}}+\dfrac{1}{{\left( {1+{a_1}} \right)\left( {1+{a_2}} \right)}}+\cdots+\dfrac{1}{{\left( {1+{a_1}} \right)\left( {1+{a_2}} \right) \cdots \left( {1+a_n} \right)}}.\end{split} \]
【难度】
【出处】
无
【标注】
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求证：$a_n<a_{n+1}$；标注答案略解析根据题意$$\begin{cases}
a_{n+1}^2+a_{n+1}-1=a_n^2 \\
{a_{n+2}}^2+{a_{n+2}}-1=a_{n+1}^2 \\
\end{cases}$$于是$$\left( {{a_{n+2}}-a_{n+1}} \right)\left( {{a_{n+2}}+a_{n+1}+1} \right)=\left( {a_{n+1}-a_n} \right)\left( {a_{n+1}+a_n} \right)$$因此 ${a_{n+2}}-a_{n+1}$ 与 $a_{n+1}-a_n$ 同正负．
而 ${a_2}^2+{a_2}-1={a_1}^2$，解得$${a_2}=\dfrac{{-1+\sqrt 5 }}{2}>0,$$于是 $a_{n+1}>a_n$． -
求证：${S_n}>n-2$；标注答案略解析$a_{n+1}-1=a_n^2-a_{n+1}^2,$ 于是$$\sum\limits_{k=1}^n {\left( {{a_k}-1} \right)}={a_1}-1+{a_1}^2-a_n^2=- \left( {1+a_n^2} \right)$$因此只需要证明 $a_n<1$，这很容易由数学归纳法证明．
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求证：${T_n}<3$．标注答案略解析设 ${b_n}=\dfrac{1}{{\left( {1+{a_1}} \right)\left( {1+{a_2}} \right) \cdots \left( {1+a_n} \right)}}$，于是
当 $n \geqslant 2$ 时，有$$\dfrac{{{b_n}}}{{{b_{n-1}}}}=\dfrac{1}{{1+a_n}}<\dfrac{1}{{1+{a_2}}}=\dfrac{1}{{1+\dfrac{{\sqrt 5-1}}{2}}}=\dfrac{{\sqrt 5-1}}{2},$$因此$$\sum\limits_{k=1}^n {{b_k}}={b_1}+\sum\limits_{k=2}^n {{b_k}}
<1+\dfrac{{\dfrac{1}{{1+\dfrac{{\sqrt 5-1}}{2}}}}}{{1-\dfrac{{\sqrt 5-1}}{2}}}
= \dfrac{{3+\sqrt 5 }}{2}<3$$因此 ${T_n}<3$ 得证．错项放缩 根据题意$$a_{n+1}^2+a_{n+1}=1+a_n^2>2a_n,$$于是$$\dfrac{1}{{1+a_{n+1}}}<\dfrac{{a_{n+1}}}{{2a_n}}.$$进而当 $n \geqslant 3$ 时，$${b_n}<\dfrac{1}{{1+{a_1}}} \cdot \dfrac{1}{{1+{a_2}}} \cdot \dfrac{{{a_3}}}{{2{a_2}}} \cdot \dfrac{{{a_4}}}{{2{a_3}}} \cdot \cdots \cdot \dfrac{{a_n}}{{2{a_{n-1}}}}=\dfrac{{a_n}}{{{2^{n-2}}}}<\dfrac{1}{{{2^{n-2}}}},$$因此$$\sum\limits_{k=1}^n {{b_k}}<{b_1}+{b_2}+\sum\limits_{k=3}^n {\dfrac{1}{{{2^{k-2}}}}}
<1+\dfrac{1}{{1+\dfrac{{\sqrt 5-1}}{2}}}+\dfrac{{\dfrac{1}{2}}}{{1-\dfrac{1}{2}}}
=\dfrac{{3+\sqrt 5 }}{2}<3.$$
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