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{2sqrt{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$left(a_{n}^{2}+1 ight)^{2}=4 a_{n}^{2} S_{n}^{2}$, $a_{n}^{2}+1=2 a_{n} S_{n}$,则$left(S_{n}-S_{n-1} ight)^{2}+1=2left(S_{n}-S_{n-1} ight) S_{n}$,于是$S_{n}^{2}-S_{n-1}^{2}=1$, $S_{n}^{2}=n$, 因此$a_{n}=sqrt{n}-sqrt{n-1}$.
(1)
egin{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}}=2left(frac{1}{sqrt{n-1}}-frac{1}{sqrt{n}} ight),]
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} 2left(frac{1}{sqrt{k-1}}-frac{1}{sqrt{k}} ight)=3-frac{2}{sqrt{n}}<3.]
类似地有
[frac{1}{S_{n}^{3}}=frac{1}{n sqrt{n}}>2left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}} ight),]
因此[sum_{k=1}^{n} frac{1}{S_{k}^{3}}=sum_{k=1}^{n} frac{1}{k sqrt{k}}>2left(1-frac{1}{sqrt{n+1}} ight)=frac{2 sqrt{n+1}-2}{sqrt{n+1}}.]