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题：求使$\small\{a_n\}\;(2a_{n+1}=a_n^2{\small-2}a_n\small+4)\,$收敛的$\small\,a_1.\;$求${\scriptsize\displaystyle\,\sum_{n\,=1}^{\infty}}\frac{1}{a_n}$$\;\overset{\text{src}}{\scriptsize(a_1:-2)}$-Pv*"x
解：令$\,h_k=a_{k+1}-a_k.\;$易见P
$(1)\quad|a_{n+1}-1|={\large\frac{|a_n-1|^2+1}{2}}.\;\;\;\therefore\;\; a_1{\small\in(0,2)\implies}|a_n{\small-1}|\small< 1$]":
$(2)\quad 2(a_{n+1}-a_n)=(a_n-2)^2.\;\;\therefore\;\;\{a_n\}\,$单调不减.]ZV
$(3)\quad a_1{\small\not\in[\,0,\,2]\implies} h_{n+1}-h_n=\frac{1}{2}{\small((a_{n+1}-2)^2-(a_n-2)^2)}$^
$\;\,\qquad\qquad\qquad\qquad\qquad\qquad=h_n(\frac{a_n+a_{n+1}}{2}-2)$:2hpM
$\;\,\qquad\qquad\qquad\qquad\qquad\qquad\ge h_n(a_2-2)$V"^
$\;\;\;\quad\qquad\qquad{\small\implies} h_{n+1}\ge h_n{\small(a_2-1)\ge\cdots\ge} h_2{\small(a_2-1)^{n-1}\to\infty}$TgH30u
$\qquad\,\big({\small a_1\not\in[0,2]\implies a_2=}\frac{(a_1-1)^2+3}{2}{\small >2\implies a_2-1>1}\big)$ISV0M
$\therefore\quad a_1\in[0,2]\iff\{a_n\}\,$收敛$\iff\displaystyle\lim_{n\to\infty}a_n=2$5L
$\qquad\color{gray}{2a_{n+1}=a_n^2-2a_n+4,\,A=\displaystyle\lim_{n\to\infty}a_n\small\implies(A-2)^2=0}$z]pV;
$\because\quad\small\dfrac{1}{a_n}-\dfrac{1}{a_n-2}=\dfrac{-2}{a_n(a_n-2)+4-4}=-\underset{\,}{\dfrac{1}{a_{n+1}-2}}$Ts
$\therefore\quad\boxed{\small\sum_{n=1}^{\infty}\frac{1}{a_n}=\frac{1}{a_1-2}\;fs
\;(a_1\not\in[\,0,\,2])}\quad\color{gray}{\small\displaystyle\sum_{n=1}^{\infty}\frac{1}{a_n}=-\frac{1}{4}\;(a_1=-2)}$k):j\#
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2020/05/31 11:05am 
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