每日一题[140] 级数不等式

数列\(\left\{a_n\right\}\)满足\(a_1=1\)，\(a_{n+1}\sqrt{\dfrac 1{a_n^2}+4}=1\)，记\(S_n=a_1^2+a_2^2+\cdots+a_n^2\)，若\(S_{2n+1}-S_n\leqslant \dfrac{m}{30}\)对任意\(n\in\mathcal N^*\)恒成立，则正整数\(m\)的最小值是________．

正确答案是\(10\)．

首先根据题意\[\dfrac{1}{a_{n+1}^2}=\dfrac{1}{a_n^2}+4,\]于是可得\[a_n^2=\dfrac{1}{4n-3},n\in\mathcal N^*.\]

进而\[\begin{split}S_{2n+1}-S_n&=a_{n+1}^2+a_{n+2}^2+\cdots+a_{2n+1}^2\\&=\dfrac{1}{4n+1}+\dfrac{1}{4n+5}+\cdots+\dfrac{1}{8n+1},\end{split}\]为了探寻其上界，先研究\(T_n=S_{2n+1}-S_n\)的单调性：\[\Delta T_n=T_n-T_{n-1}=\dfrac{1}{8n+1}+\dfrac{1}{8n-3}-\dfrac{1}{4n-3}<0,\]其中\(n\geqslant 2,n\in\mathcal N^*\)．因此其上确界为\[T_1=\dfrac{1}{5}+\dfrac{1}{9},\]从而不难求得正整数\(m\)的最小值为\(10\)．

注    事实上，注意到\[T_n=\dfrac 14\left(\dfrac{1}{n+\dfrac 14}+\dfrac{1}{n+\dfrac 54}+\cdots+\dfrac{1}{n+\left(n+\dfrac 14\right)}\right),\]于是\[\dfrac 14\int_{n+\frac 14}^{n+\frac 54}{\dfrac 1x}{\rm d}x<T_n<\dfrac 14\int_{n-\frac 34}^{2n+\frac 54}{\dfrac 1x}{\rm d}x,\]即\[\dfrac 14\ln\left(2+\dfrac{3}{4n+1}\right)<T_n<\dfrac 14\ln\left(2+\dfrac{7}{4n-3}\right),\]因此当\(n\to\infty\)时，\(T_n\to\dfrac 14\ln 2\approx 0.1732\)．