数列不等式

已知$S_n$是正项数列$\{a_n\}$的前$n$项和且$4a_n,a_n^2+1,a_nS_n^2$成等比数列.试确定数列$\{a_n\}$的通项公式.

(1)证明$a_{n+1}<\frac{\sqrt{n+1}}{2n}$;

(2)证明\[\frac{2\sqrt{n+1}-2}{\sqrt{n+1}}<\frac{1}{S_1^3}+\frac{1}{S_2^3}+\frac{1}{S_3^3}+\cdots+\frac{1}{S_n^3}<3.\]

证.由于$\left(a_{n}^{2}+1\right)^{2}=4 a_{n}^{2} S_{n}^{2}$, $a_{n}^{2}+1=2 a_{n} S_{n}$,则$\left(S_{n}-S_{n-1}\right)^{2}+1=2\left(S_{n}-S_{n-1}\right) S_{n}$,于是$S_{n}^{2}-S_{n-1}^{2}=1$, $S_{n}^{2}=n$, 因此$a_{n}=\sqrt{n}-\sqrt{n-1}$.

(1)

\begin{align*} a_{n+1}&=\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}} \\ &=\frac{\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n}) \sqrt{n+1}}<\frac{\sqrt{n+1}}{(\sqrt{n}+\sqrt{n}) \sqrt{n}}=\frac{\sqrt{n+1}}{2 n}\end{align*}

(2)注意到

\[\frac{1}{S_{n}^{3}}=\frac{1}{n \sqrt{n}}=\frac{2}{n \sqrt{n}}<\frac{2}{\sqrt{n} \sqrt{n-1}(\sqrt{n}+\sqrt{n-1})}=\frac{2(\sqrt{n}-\sqrt{n-1})}{\sqrt{n} \sqrt{n-1}}=2\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}}\right),\]

于是

\[\sum_{k=1}^{n} \frac{1}{S_{k}^{3}}=\sum_{k=1}^{n} \frac{1}{k \sqrt{k}}=1+\sum_{k=2}^{n} \frac{1}{k \sqrt{k}}<1+\sum_{k=2}^{n} 2\left(\frac{1}{\sqrt{k-1}}-\frac{1}{\sqrt{k}}\right)=3-\frac{2}{\sqrt{n}}<3.\]

类似地有

\[\frac{1}{S_{n}^{3}}=\frac{1}{n \sqrt{n}}>2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right),\]

因此\[\sum_{k=1}^{n} \frac{1}{S_{k}^{3}}=\sum_{k=1}^{n} \frac{1}{k \sqrt{k}}>2\left(1-\frac{1}{\sqrt{n+1}}\right)=\frac{2 \sqrt{n+1}-2}{\sqrt{n+1}}.\]