题目 已知数列 $\{a_n\}$ 满足 $a_1=1$，$a_{n+1}=a_n^2+a_n$． 问题1 求证：$\displaystyle\sum_{k=1}^n\dfrac{1}{a_k+1}<1$； 答案1 略 解析1 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． 备注1 问题2 求证：$\displaystyle\sum_{k=1}^n\dfrac{1}{\sqrt{a_k}+2}<1$． 答案2 略 解析2 注意到\[\sqrt{a_{n+1}}+2=\sqrt{a_n^2+a_n}+2>a_n+1,\]于是\[\begin{split} \sum_{k=1}^n\dfrac{1}{\sqrt{a_k}+2}&=\sum_{k=1}^4\dfrac{1}{\sqrt{a_k}+2}+\sum_{k=5}^n\dfrac{1}{\sqrt{a_k}+2}\\<br/>&<\sum_{k=1}^4\dfrac{1}{\sqrt{a_k}+2}+\sum_{k=4}^{n-1}\dfrac{1}{a_k+1}\\<br/>&<\sum_{k=1}^4\dfrac{1}{\sqrt{a_k}+2}+\dfrac{1}{a_4}\\<br/>&=\dfrac{1}{3}+\dfrac{1}{\sqrt 2+2}+\dfrac{1}{\sqrt{6}+2}+\dfrac{1}{\sqrt{42}+2}+\dfrac{1}{42}\\<br/>&=0.992\cdots\\<br/>&<1,\end{split}\]于是原命题得证． 备注2