已知数列 $\{a_n\}$ 满足 $a_1=1$，$a_{n+1}=a_n^2+a_n$．
【难度】
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【标注】
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求证：$\displaystyle\sum_{k=1}^n\dfrac{1}{a_k+1}<1$；标注答案略解析根据题意，有\[\dfrac{1}{a_{n+1}}=\dfrac{1}{a_n}-\dfrac{1}{a_n+1},\]于是\[\sum_{k=1}^n\dfrac{1}{a_k+1}=\dfrac{1}{a_1}-\dfrac{1}{a_{n+1}}<1,\]命题得证．
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求证：$\displaystyle\sum_{k=1}^n\dfrac{1}{\sqrt{a_k}+2}<1$．标注答案略解析考虑裂项放缩\[\dfrac{1}{\sqrt{a_{k+1}}+2}=\dfrac1{\sqrt{a_k^2+a_k}+2}<\dfrac{1}{a_k+\lambda}-\dfrac{1}{a_{k+1}+\lambda},\]设函数\[f(x)=\dfrac{1}{x+\lambda}-\dfrac{1}{x^2+x+\lambda}-\dfrac{1}{\sqrt{x^2+x}+2},\]则\[f(x)=\dfrac{x^2}{(x+\lambda)(x^2+x+\lambda)}-\dfrac{1}{\sqrt{x^2+x}+2},\]即\[
f(x)=\dfrac{1}{x+\lambda+1+2\lambda\cdot \dfrac 1x+\dfrac{\lambda^2}{x^2}}-\dfrac{1}{\sqrt{x^2+x}+2},\]取 $\lambda+1$，则当 $x\geqslant 6$ 时，有\[\sqrt{x^2+x}+2>x+\dfrac 25+2>x+2+\dfrac 2x+\dfrac{1}{x^2},\]因此就得到了当 $k\geqslant 3$ 时，有\[\dfrac{1}{\sqrt{a_{k+1}}+2}<\dfrac{1}{a_k+1}-\dfrac{1}{a_{k+1}+1},\]因此\[\begin{split} \sum_{k=1}^n\dfrac{1}{\sqrt{a_k}+2}&=\sum_{k=1}^3\dfrac{1}{\sqrt{a_k}+2}+\sum_{k=4}^n\dfrac{1}{\sqrt{a_k}+2}\\
&=\sum_{k=1}^3\dfrac{1}{\sqrt{a_k}+2}+\sum_{k=3}^n\left(\dfrac{1}{a_k+1}-\dfrac{1}{a_{k+1}+1}\right)\\
&<\dfrac 13+\dfrac{1}{\sqrt 2+2}+\dfrac{1}{\sqrt 6+2}+\dfrac{1}{6+1}\\
&=\dfrac{10}{21}+\dfrac{\sqrt 6-\sqrt 2}2\\
&<1,\end{split}\]原命题得证．
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