级数收敛

设$a_n=\displaystyle \sum_{k=1}^n\dfrac 1k-\ln n$.

(1) 求证:$\displaystyle \lim_{n\to +\infty}a_n$存在;

(2) 记$\displaystyle \lim_{n\to +\infty}a_n=C$,讨论级数$\displaystyle \sum_{n=1}^{+\infty}\left(a_n-C\right)$的敛散性.


分析与解 (1) 我们熟知当$x>0$且$x\ne 1$时,有$\ln x< x-1$,将$x=\dfrac n{n+1}$代入,有\[-\ln\dfrac{n+1}{n}<-\dfrac{1}{n+1},\]于是\[a_{n+1}-a_n=\dfrac{1}{n+1}-\ln\dfrac{n+1}{n}<0,\]因此$\{a_n\}$单调递减.又\[\sum_{k=1}^n\dfrac 1k=\sum_{k=1}^{n-1}\dfrac 1k+\dfrac 1n\geqslant \int_{1}^n\dfrac 1x{\rm d} x+\dfrac 12\left(1-\dfrac 1n\right)+\dfrac 1n=\ln n+\dfrac{n+1}{2n}>\ln n+\dfrac 12,\]于是$\{a_n\}$有下界$\dfrac 12$.于是$\displaystyle \lim_{n\to +\infty}a_n$存在.

(2) 考虑到\[a_n=1+\sum_{k=2}^n\left(\dfrac 1k-\ln\dfrac{k}{k-1}\right),\]而该级数收敛于$C$,且$a_n-C>0$,考虑\[a_n-C=\sum_{k=n+1}^{+\infty}\left(\ln\dfrac{k}{k-1}-\dfrac 1k\right).\]我们熟知,当$x>1$时,有$\ln x>\dfrac{2(x-1)}{x+1}$,令$x=\dfrac{k}{k-1}$,则有\[\ln\dfrac{k}{k-1}>\dfrac{2}{2k-1},\]于是\[a_n-C>\sum_{k=n+1}^{+\infty}\left(\dfrac{2}{2k-1}-\dfrac 1k\right)=\sum_{k=n+1}^{+\infty}\dfrac{1}{k(2k-1)}>\dfrac 12\sum_{k=n+1}^{+\infty}\left(\dfrac 1{k-\dfrac 12}-\dfrac 1{k+\dfrac 12}\right)=\dfrac{1}{2n+1}.\]由于级数$\displaystyle \sum_{n=1}^{+\infty}\dfrac{1}{2n+1}$发散,所以级数$\displaystyle \sum_{n=1}^{+\infty}\left(a_n-C\right)$发散.

 有趣的是,考虑到\[b_n=a_n-\dfrac{1}{2n},\]则\[\lim_{n\to+\infty}b_n=\lim_{n\to+\infty}a_n=C.\]而\[b_{n+1}-b_n=\dfrac{1}{n+1}-\ln\dfrac{n+1}{n}+\dfrac{1}{2n(n+1)}=\dfrac{2n+1}{2n(n+1)}-\ln\left(1+\dfrac 1n\right).\]考虑函数\[\varphi(x)=\ln x-\dfrac{x^2-1}{2x},\]在$x>1$时,有$\varphi(x)<0$.令$x=1+\dfrac 1n$,于是就得到了\[b_{n+1}-b_n>0,\]进而$\{b_n\}$单调递增.

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