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$nin N^+,a_n<5$
2)证明:不存在$Mle4$,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$a_n<5(nin N^+)$
2)由(1)$dfrac{1}{a_n}ge dfrac{1}{n}+dfrac{1}{5},a_n<dfrac{5n}{n+5},(nge3)$
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$a_ngedfrac{(n+1)(n+5)}{n^2+6n+10}a_{n+1}$
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$dfrac{1}{a_n}gedfrac{1}{a_{n+1}}+dfrac{n+5}{n^3+6n^2+10n}a_na_{n+1}
gedfrac{1}{a_{n+1}}+dfrac{17}{20n(n+1)},(nge3)$
累加得$dfrac{1}{a_3}gedfrac{1}{a_n}+dfrac{17}{20}(dfrac{1}{3}-dfrac{1}{n})$
代入$a_3=dfrac{15}{8}$得,$a_ngedfrac{20n}{5n+17}
ightarrow 4$
故不存在$Mle4$,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。