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题：设$\;a_1 > 0,\; a_{n+1} = \log(1+a_n),\;$求$\;{\displaystyle\underset{\,}{\lim_{n\to\infty}}}\frac{\large n(na_n-2)}{\log n}.\quad$ srcA
解：易见$\underset{\,}{\,}0< a_{n+1}< a_n\to\small A=\log(1+A)\;$故$\;a_n{\scriptsize\searrow}\,\small A=0\;\small(n\to\infty).$.
定义差分$\;\;\Delta[f](n)=f(n+1)-f(n),\;\Delta^{n+1}f=\Delta(\Delta^n f)\tiny\underset{\,}{.}$d`
$\underset{\,}{\because}\;\Delta(a_n^{-1})=-\frac{\Delta a_n}{a_n a_{n+1}}=\frac{\frac{1}{2}a_n^2+O(a_n^3)}{a_na_{n+1}}\sim \frac{1}{2\large\frac{1}{a_n}\log(1+a_n)}\to \frac{1}{2}\;\small(n\to\infty)$Q
$\displaystyle{(1)\;\underset{\,}{\lim_{n\to\infty}}(na_n-2)=0:\;\;\lim_{n\to\infty} na_n =\lim_{n\to\infty}{\small\frac{n}{a_n^{-1}}}=\lim_{n\to\infty}{\scriptsize\frac{\Delta n}{\Delta a_n^{-1}}}=2}.$g
$\underset{\,}{\quad}$令$\,\tau_n=\frac{na_n-2}{a_n}\,\small(< n),\;$则$\,\;\frac{2}{n-\tau_n}=a_n\to 0,\,\therefore\; n-\tau_n\to\infty\;$并且&?yWPb
$\underset{\,}{\quad}\Delta\tau_n =\Delta(n-\frac{2}{a_n})=1+\frac{2(\log(1+a_n)-a_n)}{a_n\log(1+a_n)}=\frac{1}{6}a_n-\frac{1}{12}U(a_n^2)$45
$\underset{\,}{(\dagger)}\,\exists m:\frac{u}{n}\overset{(1)}{<} ua_n{\small <\Delta}\tau_n,\;\tau_n>\tau_m{\scriptsize +\displaystyle{\sum_{k=m}^{n-1}}}{\large\frac{u}{k}}\sim u\small\log(n)\;(n\ge m,\,6u< {\scriptsize 1})$t!N,ZT
$\displaystyle{(2)\;\underset{\,}{\lim_{n\to\infty}}{\small\frac{1}{n(na_n-2)}}=\lim_{n\to\infty}{\small\frac{a_n/2}{na_n-2}}\overset{(\dagger)}{=}\lim_{n\to\infty}\tau_n^{-1}=0.}\quad\small(\tau_{n+m}{\scriptsize\nearrow}\infty)$&
$\underset{\,}{\quad}$令$\;\delta_n= na_n-2=a_n\tau_n,\;$因$\,\tau_{m+n}{\scriptsize\nearrow\,+}\infty,\,n\,$充分大时$\,\delta_n>0.$Lo-f
$\underset{\,}{\quad}$不难验证$\;\Delta\delta_n=\Delta(na_n)=-\frac{1}{2}\delta_nU(a_n).\,(\delta_n\,$最终递减趋于$\,0).$Z,B<
$\underset{\,}{\quad}$据$\,(\dagger)\,\;\Delta \tau_n < \frac{a_n}{2},\;{\small\bigg|\frac{\large\Delta \tau_n}{\large\Delta \delta_n^{-1}}\bigg|< \bigg|\frac{\large \delta_n\delta_{n+1}a_n}{2\large\Delta \delta_n}\bigg|=U(\delta_{n+1})}\to 0\;(n\to\infty)$*c
$\displaystyle{(3)\;\underset{\,}{\lim_{n\to\infty}}n(na_n-2)^2=0:\,\lim_{n\to\infty}{\small\frac{(na_n-2)^2}{a_n}}=\lim_{n\to\infty}{\small\frac{\tau_n}{\delta_n^{-1}}}=0.}$l
$\underset{\,}{\quad}$最后由$\;\Delta(n\delta_n)=\small\Delta(n^2a_n-2n)=(2n+1)a_n+(n+1)^2\Delta a_n-2$:i
$\underset{\,}{\qquad\qquad} = (2n+1)a_n+(n+1)^2\big(-\frac{a_n^2}{2}+\frac{a_n^3}{3}+O(a_n^4)\big)-2$d
$\underset{\,}{\qquad\qquad} = \frac{1}{3}a_n(n^2a_n^2-3na_n+3)-\frac{1}{2}(na_n-2)^2+O(a_n^2)$U=Unl
$\underset{\,}{\qquad\qquad} =\frac{1}{3}a_nU(1)+\delta_n^2O(1)$ 2E
$\quad$及$\,(3)\,$得-o
$\displaystyle{\quad\lim_{n\to\infty}{\small\frac{n(na_n-2)}{\log n}}{\small=}\lim_{n\to\infty}{\small\frac{n\Delta(n\delta_n)}{n\Delta(\log n)}}{\small=}\lim_{n\to\infty}{\small\frac{na_nU(1)+n\delta_n^2O(1)}{3\log(1+\frac{1}{n})^n}}\small=\frac{2}{3}}.$$<a.5(
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2017/09/29 03:42am 
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