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# 146 Class Note for MATH 403 at PSU

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This 4 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Pennsylvania State University taught by a professor in Fall. Since its upload, it has received 26 views.

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Date Created: 02/06/15
Math 403 Lecture Notes for Week 2 1 Limsup and Liminf De nition 1 Let A Q R If A has an upper bound7 we let supA be the least upper bound of A When A is unbounded we sometimes write supA 00 If A has a lower bound7 we let ian be the greatest lower bound of A7 and write ian 700 when A has no lower bound For a sequence ltfngtn21 we let supngt1 In supIn n 2 l and similarly for inf De nition 2 For a sequence ltfngtn21 we de ne the limit superior limsup to be limsup lim In inf sup In ngtoo new 7121107 We de ne the limit inferior to be liminf lim In sup inf In 400 naoo n21 an Example 3 1 Consider the sequence with In We compute 7 1 1 lim In inf supi inf 7 0 naoo n21n2n k n21 n 1 limI supinf7sup00 nix 21an k 721 Note that these both are equal to the limit of the sequence 2 Consider the sequence 17 711 711 H We have T f f11 1m In 1n supIn 21 ngtoo n21ngtn lim In sup inf In sup 71 71 ngtoo 1an n21 If a sequence is bounded both above and below then the limsup and liminf both exist as real numbers We can think of the limsup as the largest number which the sequence approaches in nitely often7 and limsup the smallest Proposition 4 lim In g End In inaoo Proof Set 39 f 2n sup In kgn so that lim inf In sup yn n21 l39 39 f 1m sup In 1 2n Note that yn S 2n for all n and if n S m then yn S ym and 2m S zni From this we see that yn S 2m for any two numbers n and m so that lim inf In S lim sup In D Note 5 This proposition is also true for unbounded sequences with the con vention that 700 S 700 lt 7 lt 00 S 00 for any real number Ti Proposition 6 For a sequence ltfngtn21 we have limnH00 In y and only if limn oo In limnH00 In y Proof Suppose lim In y we will show that liimn oo In End In y We consider lim sup rst For any 6 gt 0 there is an N such that lIn 7 yl lt e for all n 2 Ni Hence sukaN 1k S y 6 so limsupIn inf supk 2 nIk S ye n21 Since this is true for any 6 gt 0 we have limsupI S y Similarly we nd lim inf In 2 yr Since we know liminf In S lim sup In we conclude that they both equal yr For the other direction suppose liimn oo In End In y we will show that limIn yr Suppose not so there is some 6 gt 0 so that for all N there is an n 2 n with lIn 7 yl 2 6 ie either In 2 y e or In S y 7 6 Hence there are in nitely many n where this happens so there are either in nitely many n for which In 2 y e or there are in nitely many n for which In S y 7 e or both In the rst case we see that limsup In 2 y e and in the second case we have lim inf In S y 7 e contradicting that lim sup In lim inf In y D Proposition 7 For a sequence ltfngtn21 lim inf In 7 lim sup 7In Proof We compute limian su inf I n n2 an k sup infIk k 2 n n21 sup 7 sup7Ik k 2 n from homework n21 739f 7 zkgt iglsup 1k Sn 7 inf su 7In nzlkga 7 lim sup 7In 2 Properties of Limits Proposition 8 Let Ingtn21zmd ltyngtn21 be sequences with limIn a and lim yn I Then 1 limInyn ab 21imcIncaforc lR 5 limInynab Proof 1 Let 6 gt 0 be given We then know that there is an N1 such that lIn 7 al lt 3 for all n 2 N1 and an N2 such that lyn 7 bl lt 5 for all n 2 N2 Letting N maxN1N2 we have yn 7 a bl S lIn 7 al lyn 7 bl lt 3 3 e for all n 2 N so the de nition of limit is satis ed 2 This is a special case of 3 when yn c for all mi 3 Assume for simplicity that a gt 0 and b gt 0 the other cases are similar Let 6 gt 0 be given Note that lznyn7abl lrnyn7znbznb7abl S lznyn7bllbzn7al We choose N1 such that lIn 7 al lt i for n 2 Nli We choose N2 so that lIn 7 al lt for n 2 N2 iiei lt In lt gal We then choose N3 so that lyn 7 bl lt 37 for n 2 Ngi Then letting N maxN1N2N3 we have that for n 2 N 3 lInyn7ablS7aieri e 2 a 3 2b and hence the limit de nition is satis ed D Note 9 We must be careful when dealing with in nite limitsi When lim In 00 and limyn 700 the sequence ltIn yngt may or may not have a limit Similarly if lim In 00 and lim yn 0 the sequence ltIn may or may not have a limit Also the converses of the above can fail For instance if In n and yn 7n then limIn yn 0 but limIn and limyn do not exist as real numbers Similarly when In n and yn i we have limIn yn 1 but lim In 00 and limyn 0 Theorem 10 Comparison Theorem Let ltfngtn21 and yngtn21 be se quences such that In S yn for all n and suppose the limits of both sequences eIz39st Then lim In S lim yn Proof We note that supkgtn 1k S supkgtnyk for all n7 since 1k S yk for all kl Hence 39 f lt 39 f 351212 7 sleaze so that MIn S Eynl Since these are equal to the respective limits7 the theorem follows E The same is true for lim sup and liminfl Note that we do not always have strict inequality The sequence with In 0 for all n and yn satis es In lt yn for all n7 but the limits are both equal to 0 Theorem 11 Squeeze Theorem Let ltfngtn21 ltyngtn2h and lt2ngtn21 be sequences with In S yn S 2n for all n Suppose limIn lim 2n Then lim yn lim In lim 2n Proof Let limIn lim 2n ll Let 6 gt 0 be given We can choose an N large enough so that lIn 71 lt e and lzn 71 lt 6 whenever n 2 Ni Then 7eltIn7lSyn7lSzn7llte so lyn 7 ll lt 6 whenever n 2 N7 satisfying the limit de nition D Example 12 We can use the squeeze theorem to compute the limit of the sequence with yn by letting In 7 2n i and noting that In S yn S 2n and that limIn lim 2n 0i

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