MT【131】$a_{n+1}cdot a_n=dfrac 1n$

已知数列({a_n})满足(a_1=1)，(a_{n+1}cdot a_n=dfrac 1n)((ninmathbb N^*))．
(1) 求证：(dfrac{a_{n+2}}{n}=dfrac{a_n}{n+1})；
(2) 求证：(2left(sqrt{n+1}-1 ight)leqslant dfrac{1}{2a_3}+dfrac{1}{3a_4}+cdots+dfrac{1}{(n+1)a_{n+2}}leqslant n)．

解:(1) 根据题意，有$$egin{split} dfrac{a_{n+2}}{n}=&dfrac{dfrac{1}{n+1}cdot dfrac{1}{a_{n+1}}}{n}=&dfrac{na_n}{n(n+1)}=dfrac{a_n}{n+1}.end{split}$$
(2) 根据第((1))小题的结论，有(dfrac{1}{2a_3}+dfrac{1}{3a_4}+cdots+dfrac{1}{(n+1)a_{n+2}}=a_2+a_3+cdots+a_{n+1}.)
右边不等式　根据第((1))小题的结论，有(dfrac{a_{n+2}}{a_n}=dfrac{n}{n+1}<1,)于是数列的奇子列和偶子列均单调递减，结合(a_1=a_2=1)，可得(a_nleqslant 1,ninmathbb N^*)，于是右边不等式得证．
左边不等式　由于(egin{split}dfrac{1}{a_ncdot a_{n+1}}&=n,dfrac{1}{a_{n+1}cdot a_{n+2}}&=n+1,end{split})于是(dfrac{1}{a_{n+1}}left(dfrac{1}{a_{n+2}}-dfrac{1}{a_n} ight)=1,)从而(a_{n+1}=dfrac{1}{a_{n+2}}-dfrac{1}{a_n}.) 因此$$egin{split} a_2+a_3+cdots+a_{n+1}=&dfrac{1}{a_{n+2}}+dfrac{1}{a_{n+1}}-dfrac{1}{a_1}-dfrac{1}{a_2}geqslant &dfrac{2}{sqrt{a_{n+1}a_{n+2}}}-2=&2left(sqrt{n+1}-1 ight),end{split} $$于是左边不等式得证．
综上所述，原命题得证．
评:这类题目最后往往要用基本不等式把(a_{n+1}cdot a_n=dfrac 1n)这个条件用进去.下面给一道类似的练习:
设(a_1=1,a_ncdot a_{n+1}=n,nin N^+),求证:(sumlimits_{k=1}^{n}{dfrac{1}{a_k}}ge2sqrt{n}-1)