MT【198】连乘积放缩

(2018中科大自招最后一题)
设$a_1=1,a_{n+1}=\left(1+\dfrac{1}{n}\right)^3(n+a_n)$证明:
(1)$a_n=n^3\left(1+\sum\limits_{k=1}^{n-1}\dfrac{1}{k^2}\right);
(2)\prod\limits_{k=1}^n\left(1+\dfrac{k}{a_k}\right)<3$

证明:
1)数学归纳法，略.

$k=1$时候显然成立,$k\ge2$时有如下漂亮的连乘积放缩：

\begin{align*}
\prod\limits_{k=1}^n\left(1+\dfrac{k}{a_k}\right)&=\prod\limits_{k=1}^n\left(1+\dfrac{1}{k^2(1+\sum\limits_{m=1}^{k-1}\frac{1}{m^2})}\right)\\
&<\prod\limits_{k=1}^n(1+\dfrac{1}{k^2\left(2-\frac{1}{k}\right)})\\
&=\prod\limits_{k=1}^{n}{\dfrac{2k^2-k+1}{2k^2-k}}\\
&<2\prod_{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
\begin{align*}
\sum\limits_{k=1}^n\dfrac{k}{a_k}& =\sum\limits_{k=1}^n\dfrac{k}{k^3\left(1+\sum\limits_{m=1}^{k-1}\frac{1}{m^2}\right)} \\
& <\sum\limits_{k=1}^n\dfrac{1}{k^2\left(2-\frac{1}{k}\right)}\\
&=\sum\limits_{k=1}^n\dfrac{1}{2k^2-k}\\
&<\sum\limits_{k=1}^n\dfrac{1}{2(k-\frac{3}{4})(k+\frac{1}{4})}\\
&=2-\dfrac{1}{2n+1/2}\\
&<2
\end{align*}
故
\begin{align*}
\prod\limits_{k=1}^n\left(1+\dfrac{k}{a_k}\right)& \le\left(\dfrac{\sum\limits_{k=1}^n{(1+\dfrac{k}{a_k}})}{n}\right)^n \\
& <\left(1+\dfrac{2}{n}\right)^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：

$\lim\limits_{n\longrightarrow +\infty}{(1+\dfrac{1}{n})^n}=e$

练习:证明存在:$n\in N,\prod\limits_{k=1}^n\left(\dfrac{k^2}{k^2+1}\right)<\dfrac{2}{7}$