(2018中科大自招最后一题)
设$a_1=1,a_{n+1}=left(1+dfrac{1}{n}
ight)^3(n+a_n)$证明:
(1)$a_n=n^3left(1+sumlimits_{k=1}^{n-1}dfrac{1}{k^2}
ight);
(2)prodlimits_{k=1}^nleft(1+dfrac{k}{a_k}
ight)<3$
证明:
1)数学归纳法，略.
$k=1$时候显然成立,$kge2$时有如下漂亮的连乘积放缩：
egin{align*}
prodlimits_{k=1}^nleft(1+dfrac{k}{a_k}
ight)&=prodlimits_{k=1}^nleft(1+dfrac{1}{k^2(1+sumlimits_{m=1}^{k-1}frac{1}{m^2})}
ight)\
&<prodlimits_{k=1}^n(1+dfrac{1}{k^2left(2-frac{1}{k}
ight)})\
&=prodlimits_{k=1}^{n}{dfrac{2k^2-k+1}{2k^2-k}}\
&<2prod_{k=2}^{n}{dfrac{k(2k-1)}{(k-1)(2k+1)}}\
&=dfrac{6n}{2n+1}\
&<3
end{align*}
如果证明$<8$则变为一道难度降为高考题的题，可以解答如下：
由于
egin{align*}
sumlimits_{k=1}^ndfrac{k}{a_k}& =sumlimits_{k=1}^ndfrac{k}{k^3left(1+sumlimits_{m=1}^{k-1}frac{1}{m^2}
ight)} \
& <sumlimits_{k=1}^ndfrac{1}{k^2left(2-frac{1}{k}
ight)}\
&=sumlimits_{k=1}^ndfrac{1}{2k^2-k}\
&<sumlimits_{k=1}^ndfrac{1}{2(k-frac{3}{4})(k+frac{1}{4})}\
&=2-dfrac{1}{2n+1/2}\
&<2
end{align*}
故
egin{align*}
prodlimits_{k=1}^nleft(1+dfrac{k}{a_k}
ight)& leleft(dfrac{sumlimits_{k=1}^n{(1+dfrac{k}{a_k}})}{n}
ight)^n \
& <left(1+dfrac{2}{n}
ight)^n\
&<e^2<8
end{align*}
改为$<8$后本质上考察了下面这个重要的极限：
$limlimits_{nlongrightarrow +infty}{(1+dfrac{1}{n})^n}=e$
练习:证明存在:$nin N,prodlimits_{k=1}^nleft(dfrac{k^2}{k^2+1} ight)<dfrac{2}{7}$