MT167反复放缩

Posted 2020-11-01 青春的记忆

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,$M>a_n$

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。