MT【167】反复放缩

已知数列$\{a_n\}$满足:$a_1=1,a_{n+1}=a_n+\dfrac{a_n^2}{n(n+1)}$
1)证明:对任意$n\in N^+,a_n<5$
2)证明:不存在$M\le4$,使得对任意$n,a_n<M$

证明:
1)显然$a_{n+1}>a_n,a_{n+1}=a_n+\dfrac{a_n^2}{n(n+1)}<a_n+\dfrac{a_na_{n+1}}{n(n+1)}$
故$\dfrac{1}{a_n}<\dfrac{1}{a_{n+1}}+\dfrac{1}{n(n+1)}$ 累加得:$\dfrac{1}{a_3}<\dfrac{1}{a_n}+\dfrac{1}{3}-\dfrac{1}{n}$
由于$a_1=1,a_2=\dfrac{3}{2},a_3=\dfrac{15}{8}$代入上式得$\dfrac{1}{a_n}\ge \dfrac{1}{n}+\dfrac{1}{5}>\dfrac{1}{5}$.故$a_n<5(n\in N^+)$
2)由(1)$\dfrac{1}{a_n}\ge \dfrac{1}{n}+\dfrac{1}{5},a_n<\dfrac{5n}{n+5},(n\ge3)$
故$a_{n+1}=a_n+\dfrac{a_n^2}{n(n+1)}<a_n+\dfrac{\frac{5n}{n+5}a_n}{n(n+1)}=\dfrac{n^2+6n+10}{(n+1)(n+5)}a_n$
故$a_n\ge\dfrac{(n+1)(n+5)}{n^2+6n+10}a_{n+1}$
故$a_{n+1}=a_n+\dfrac{a_n^2}{n(n+1)}\ge a_n+\dfrac{\frac{(n+1)(n+5)}{n^2+6n+10}a_na_{n+1}}{n(n+1)}=a_n+\dfrac{n+5}{n^3+6n^2+10n}a_na_{n+1}$
故$\dfrac{1}{a_n}\ge\dfrac{1}{a_{n+1}}+\dfrac{n+5}{n^3+6n^2+10n}a_na_{n+1}
\ge\dfrac{1}{a_{n+1}}+\dfrac{17}{20n(n+1)},(n\ge3)$
累加得$\dfrac{1}{a_3}\ge\dfrac{1}{a_n}+\dfrac{17}{20}(\dfrac{1}{3}-\dfrac{1}{n})$
代入$a_3=\dfrac{15}{8}$得,$a_n\ge\dfrac{20n}{5n+17}\rightarrow 4$
故不存在$M\le4$,使得对任意$n,a_n<M$

注:此类题型也较常见，但往往最后一步裂项放缩要观察一下。