每日一题[988]数列与数列极限

已知$n$是正整数，数列$\{a_k\}$满足$a_1=\dfrac{1}{n(n+1)}$，且\[a_{k+1}=-\dfrac{1}{k+n+1}+\dfrac nk\sum_{i=1}^ka_i,\]其中$k=1,2,\cdots$．

(1) 求$a_2,a_3$；
(2) 求数列$\{a_k\}$的通项；
(3) 设$b_n=\displaystyle \sum_{k=1}^n\sqrt{a_k}$，求证：$\lim\limits_{n\to \infty}b_n=\ln 2$．

分析与解　(1) 根据已知，有\[a_2=-\dfrac{1}{n+2}+n\cdot \dfrac{1}{n(n+1)}=\dfrac{1}{(n+1)(n+2)},\]且\[a_3=-\dfrac{1}{n+3}+\dfrac n2\left(\dfrac{1}{n(n+1)}+\dfrac{1}{(n+1)(n+2)}\right)=\dfrac{1}{(n+2)(n+3)}.\]
(2) 由$a_1,a_2$容易归纳出$$a_k=\dfrac{1}{(n+k-1)(n+k)},k\in\mathbb N^*.$$下面用数学归纳法证明：

当$k=1$时，由已知知结论成立；

若对$i\leqslant k$，有$$a_i=\dfrac {1}{(n+i-1)(n+i)}$$成立；则对$i=k+1$，有\[\begin{split} a_{k+1}=&-\dfrac{1}{k+n+1}+\dfrac nk\sum_{i=1}^k\dfrac {1}{(n+i-1)(n+i)}\\=&-\dfrac{1}{k+n+1}+\dfrac nk\sum_{i=1}^k\left(\dfrac 1{n+i-1}-\dfrac 1{n+i}\right)\\=&-\dfrac{1}{k+n+1}+\dfrac nk\left(\dfrac 1n-\dfrac 1{n+k}\right)\\=&-\dfrac{1}{k+n+1}+\dfrac nk\cdot\dfrac {k}{n(n+k)}\\=&\dfrac{1}{(n+k)(n+k+1)}.\end{split}\]所以对$i=k+1$命题也成立；

由数学归纳法知，对任意$k\in\mathbb{N}^*$，有$a_k=\dfrac{1}{(n+k-1)(n+k)}$．

(3) 因为$$\dfrac 1{n+k}<\sqrt{a_k}<\dfrac 1{n+k-1},$$所以有\[\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}<b_n<\dfrac 1n+\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n-1}.\]考虑左边，有\[\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n-1}>\int_{n+1}^{2n}\dfrac 1x{\rm d} x=\ln\dfrac{2n}{n+1}.\]类似的，对于右边，有\[\dfrac{1}{n}+\dfrac{1}{n+1}+\cdots+\dfrac{1}{2n-1}<\int_{n-1}^{2n-1}\dfrac 1x{\rm d}x=\ln \dfrac{2n-1}{n-1}.\]于是有\[\ln\dfrac{2n}{n+1}<b_n<\ln\dfrac{2n-1}{n-1},\]因此\[\lim\limits_{n\to \infty}b_n=\ln 2.\]