浅谈数学分析中有关数列极限的几种求法 2

《数列极限几种解法初探》湛江师范学院数科院09数本2梁德君20092245011数列极限几种求法初探梁德君（湛江师范学院数学与计算科学学院湛江524048）摘要：极限论是数学分析的基础，极限的问题一直是数学分析的困难之一，也是许多科学领域的重要思想之一，然而数列极限又是极限的基础。同时涉及数列极限的问题有很多，包括极限的求法，数列极限的证明，极限的存在等。因为极限的重要性，从而怎样求极限也显得尤其重要。一般极限还可以用定义来求，但对于一些复杂极限，直接按照极限的定义求就显得非常困难，不仅计算量大，而且不一定能求出最终结果，下面初步总结了极限的几种求法，并且以实例来加以说明。关键词：数列极限，定义法，级数性法，迫敛性，单调有界原理，压缩法，柯西判别法，错位法，拆分法，公式法，定积分定义法，定积分性质法，归结原则PreliminaryanalysisabouttheseveralmethodsofSeriesLimitLIANGDejun(Mathematicsandcomputationalscienceschool,ZhanjiangNormalUniversity,Zhanjiang,524048)Abstract:Thelimitisthebasisofmathematicalanalysis,limitquestionisalwaysoneofthedifficultiesofmathematicalanalysis,anditisalsooneoftheimportantideasinmanyscientificfields.However,Serieslimitisthebasisofthelimit.Atthesametimetheproblemswhatrelatetoserieslimitareolot,includingtothemethodsoflimit,thepoorofserieslimit,theexistenceofsuchlimits.Becauseoftheimportanceoflimits,itisparticularlyimportanthowitisdone.Definitioncanbeusedtosolvegenerallimit,butitisverydifficultforanumberofcomplexlimit.Notonlyittakesusmuchtimetosolveit,butalsowenotsuretogettheresults.Belowseveralmethodswillbedescribedandseveralexampleswillbeusedtoillustratethem.Keyword：Series-LimitDefinition-methodSeriesresistancemethodconvergence-propertymonotonicity-principlecompression-processCauchy'sconvergencetestformula-methodDefiniteintegraldefinitionmethodDefiniteintegralpropertymethodPrinciplesboildown《数列极限几种解法初探》湛江师范学院数科院09数本2梁德君20092245012目录1、最基本最常用的方法—定义法····························32、迫敛性················································33、判别数列极限的几种方法································53.1柯西收敛准则········································53.2单调有界原理·······································53.3两种判别方法·······································73.3.1有界变差数列···································73.3.2压缩数列·······································74、利用函数的某些性质来求数列极限······················75、利用施笃兹公式以及其它方法求解······················95.1施笃兹公式求解·································95.2级数求解········································105.3定积分定义法······································125.4定积分性质法······································125.5错位法············································135.6拆分法············································136、结语················································147、参考文献············································14《数列极限几种解法初探》湛江师范学院数科院09数本2梁德君200922450131、最基本最常用的方法—定义法定义1设na为数列,a为实数,若对任给的正数,总存在正整数N,使得当nN时有||naa,则称数列na收敛于a,实数a称为数列na的极限,并记作limnnaa或()naan.读作：当n趋于无穷大时,na的极限等于a或na趋于a。由于n限于取正整数,所以在数列极限的记号中把n写成n,即limnnaa或()naan.例1、设lim0nnxaxax。证明limnxxa。证明方法（1）：因为lim0nnxaxax。故0（取21），NN,Nn,有nxa22.nnnxaxaaxaa于是241naxaa，由的任意性limnxxa。证明方法（2）：2limlim(1)0nxxnnxaaxaxa2lim1,limnxxnaxaxa从而小结：设通过此例总结出运用”“N-论证法的大致步骤：１）任意给定0；２）令Axn；３）推出)(n；４）取)(N，再用N-语言顺述并得出结论。2、迫敛性定理2.1（迫敛性）设收敛数列na，nb都以a为极限，数列nc满足存在正整数0N，当nNn时有，nnncba，则数列nc收敛，且acnnlim。例2、求极限21lim(1)sinnnkkknn。《数列极限几种解法初探》湛江师范学院数科院09数本2梁德君20092245014二、解法1容易发现直接化为黎曼和的形式有困难.注意到3sin()xxOx=+,由于33336611|1()|20,()nnkkkkkOCnnnn，所以2211lim1sinlim1nnnnkkkkkknnnn65)(1)(lim102122dxxxnnknknkn.解法2（利用数列夹逼准则）利用31sin6xxxx，得3326221sin6kkkknnnn332622111111(1)1sin16nnnnkkkkkkkkkkkknnnnnnnn,由于33336611|1|20,()nnkkkkknnnn，21lim1nnkkknn65)(1)(lim102122dxxxnnknknkn，所以215lim(1)sin6nnkkknn.思考题（华南师大2000年考研真题）求证21)22211(lim222nnnnnnnx。小结：运用此法的关键将适当放大与缩小，一般是从数列na出发，将其通项放大后得数列，缩小后得数列nb，并使na与nb的极限都存在且相等，放缩的技巧基本上类似应用N-定义证数列极限时的常用方法，关键是掌握不等式的放缩的各种方法。但大多数数列并不是有一定规律的或很容易使用迫敛性就可以求之的，而且有的数列是有极限还得进行判断，这时就得引入判别数列极限存在的定理。《数列极限几种解法初探》湛江师范学院数科院09数本2梁德君200922450153、判别数列极限存在的几种方法定理3.1（柯西收敛准则）：数列收敛的充分必要条件是任给0,存在)(N,使得当NNnm，时，都有nmaa成立。例3、证明发散ncos。证法一：取01cos220，Nn，取，120Nn，12m0N则2200000000sinsin2coscos,,mnmnmnNmn而（差化积）=2)1(2)1(22)1(2)1(20000sinsin2nmNNnmNN=1sin1sin2sinsin222220000nmnm=021cos2.方法二（反证法）假设收敛，ncos记nAncoslim，故.)2cos(limAnn又因nncos)2cos(=).1sin(2n于是0)1(sinlimnBn。故。101)1(sin1lim22nAn而1cos)1cos(2cos)2(cosnnn，于是上式两边取极限得1cos22AA，得1,0AA这与矛盾。从而可证数列发散ncos。定理3.2（单调有界原理）在实数系中，有界的单调数列必有极限。例4、设11x，11nnnxxx，(1,2,...)n，试证：{}nx收敛，并求limnnx。证明令xxxf1)(，则有2)1(1)(xxf，)(f，)(xf在),0(上是严格递减的；当x时，)(xf；当x时，《数列极限几种解法初探》湛江师范学院数科院09数本2梁德君20092245016)(xf；若1x，则有显然12nx，nx2，),2,1(n；将11nnnxxx代入1211nnnxxx，得22(1)(1)2nnnxxx，由nnnnnxxxxx2)1()1(22nnxx2)1()(22，得}{12nx单调递减，}{2nx单调递增，设axnn12lim，bxnn2lim，在121221nnnxxx，nnnxxx22121中，令n取极限，得aab1，bba1，从而有ba，故nnxlim或者注意到111111nnnnaxaxxx，我们有当nxa时，1111(1)1naxaaa，当nxa时，1111(1)1naxaaa，于是11xa，知21nxa，21nxa，(1,2,...)n往证21{}nx递减，2{}nx递增，实际上从11kkkaxxx中，解出111kkkaxxx1121111kkkkkkaxaxxxxx1111