每日一题[836]级数不等式的证明

数列$\{a_n\}$满足$a_1=1$,$na_{n+1}=(n+2)a_n+n$,$b_n=\dfrac{a_n}{n(n+1)}$.

(1) 求$b_n$和$a_n$;

(2) 求证:$\dfrac n2\leqslant \dfrac{1}{a_2}+\dfrac{1}{a_3}+\cdots+\dfrac{1}{a_{n+1}}+b_1^2+b_2^2+\cdots+b_n^2\leqslant \dfrac{2n^2-4n+4\sqrt n-1}{2n}$.


cover

分析与解 (1)根据题意,有\[\dfrac{a_{n+1}}{(n+1)(n+2)}=\dfrac{a_n}{n(n+1)}+\dfrac{1}{(n+1)(n+2)},\]于是\[b_{n+1}+\dfrac{1}{n+2}=b_n+\dfrac{1}{n+1},\]进而可得\[b_n+\dfrac{1}{n+1}=b_1+\dfrac{1}{2}=1,\]因此\[b_n=\dfrac{n}{n+1},a_n=n^2.\](2)对于左边不等式,考虑到\[\dfrac 1{a_{k+1}}+b_k^2=\dfrac{1}{(k+1)^2}+\dfrac{k^2}{(k+1)^2}=\dfrac{k^2+1}{(k+1)^2}\geqslant \dfrac 12,\]等号当且仅当$k=1$时取得.取$k=1,2,\cdots,n$,累加即得左边不等式.

对于右边不等式,即\[n-\sum_{k=1}^{n}\dfrac{k^2+1}{(k+1)^2}\geqslant 2+\dfrac{1}{2n}-\dfrac{2}{\sqrt n},\]也即\[\sum_{k=1}^{n}\dfrac{2k}{(k+1)^2}\geqslant 2+\dfrac{1}{2n}-\dfrac{2}{\sqrt n}.\]考虑到\[\dfrac{2k}{(k+1)^2}=\dfrac{2}{k+\dfrac 1k+2}\geqslant \dfrac{2}{k+3},\]下面证明引理:当$k\geqslant 2$时,有\[\dfrac{2}{k+3}\geqslant \dfrac{1}{2k}-\dfrac{2}{\sqrt k}-\dfrac{1}{2(k-1)}+\dfrac{2}{\sqrt{k-1}}.\]当$k=2$时,左边为$\dfrac 25$,右边为$\dfrac {7-4\sqrt 2}4$,显然成立;

当$k\geqslant 3$时,考虑证明\[\dfrac{1}{k+3}\geqslant \dfrac 1{\sqrt{k-1}}-\dfrac{1}{\sqrt k},\]事实上,有\[\dfrac{1}{\sqrt{k-1}}-\dfrac{1}{\sqrt k}=\dfrac{1}{k\sqrt{k-1}+(k-1)\sqrt k}\leqslant \dfrac{1}{\sqrt 2k+2\sqrt 3}<\dfrac{1}{k+3},\]因此引理得证.

这样就有\[\sum_{k=1}^{n}\dfrac{2k}{(k+1)^2}\geqslant\sum_{k=1}^n\dfrac{2}{k+3}
\geqslant 2+\dfrac{1}{2n}-\dfrac{2}{\sqrt n},\]命题得证.

注1 由第一步放缩可知$\displaystyle \sum_{k=1}^{n}\dfrac{2k}{(k+1)^2}$发散,而右侧收敛于$2$,因此右侧不等式可以大幅加强.

注2 考虑\[\int \dfrac{2x}{(x+1)^2}{\rm d}x=\dfrac{2}{x+1}+2\ln(x+1),\]且函数$f(x)=\dfrac{2x}{(x+1)^2}$在$x\geqslant 2$时单调递减,且为下凸函数,因此有\[\begin{split}\sum_{k=1}^{n}\dfrac{2k}{(k+1)^2}&\geqslant \dfrac 12+\int_2^{n+1}\dfrac{2x}{(x+1)^2}{\rm d} x+\dfrac{f(2)-f(n+1)}2\\&=\dfrac 12+\dfrac{2}{n+2}+2\ln (n+2)-\dfrac 23-2\ln 3+\dfrac 29-\dfrac{n+1}{(n+2)^2}\\&=2\ln\dfrac{n+2}3+\dfrac{n+3}{(n+2)^2}+\dfrac{1}{18}.\end{split}\]

此条目发表在每日一题分类目录,贴了标签。将固定链接加入收藏夹。

发表评论