Limit sup and limit inf.

1 Limit sup and Limit order to make us understand the information more on approaches of a given realsequence an n 1 , we give two definitions, thier names are upper Limit and lower Limit . Itis fundamental but important tools in of Limit sup and Limit infDefinition Given a real sequence an n 1 ,wedefinebn sup am:m n andcn inf am:m n .Example 1 1 n n 1 0,2,0,2,.. ,sowehavebn 2andcn 0 for 1 nn n 1 1,2, 3,4,.. ,sowehavebn andcn for n n 1 1, 2, 3,.. ,sowehavebn nandcn for Given a real sequence an n 1 , and thus definebnandcnas the same ,andcn n If there is a positive integerpsuch thatbp , thenbn n there is a positive integerqsuch thatcq , thencn n bn is decreasing and cn is property 3, we can give definitions on the upper Limit and the lower Limit of a givensequence as Given a real sequence an and letbnandcnas the same as before.

2 (1) If everybn R, theninf bn:n N is called the upper Limit of an , denoted bylimn is,limn supan everybn , then we definelimn supan .(2) If everycn R, thensup cn:n N is called the lower Limit of an , denoted bylimn is,limn infan everycn , then we definelimn infan .Remark The concept of lower Limit and upper Limit first appear in the book (AnalyseAlge brique) written byCauchyin 1821. But until 1882,Paul du Bois-Reymondgave explanations on them, it becomes 1 1 n n 1 0,2,0,2,.. ,sowehavebn 2andcn 0 for allnwhich implies thatlim supan 2 and lim infan 1 nn n 1 1,2, 3,4,.. ,sowehavebn andcn for allnwhich implies thatlim supan and lim infan .Example n n 1 1, 2, 3,.. ,sowehavebn nandcn for allnwhich implies thatlim supan and lim infan.

3 Relations with convergence and divergence for upper (lower) limitTheorem Let an be a real sequence, then an converges if, and only if, the upperlimit and the lower Limit are real withlimn supan limn infan limn Let an be a real sequence, then we have(1) limn supan an has no upper bound.(2) limn supan for anyM 0, there is a positive integern0suchthat asn n0,wehavean M.(3) limn supan aif, and only if, (a) given any 0, there are infinitemany numbersnsuch thata anand (b) given any 0, there is a positive integern0such that asn n0,wehavean a .Similarly, we also haveTheorem Let an be a real sequence, then we have(1) limn infan an has no lower bound.(2) limn infan for anyM 0, there is a positive integern0suchthat asn n0,wehavean M.

4 (3) limn infan aif, and only if, (a) given any 0, there are infinitemany numbersnsuch thata anand (b) given any 0, there is a positive integern0such that asn n0,wehavean a .From Theorem 2 an Theorem 3, the sequence is divergent, we give the Let an be a real sequence, then we have(1) If limn supan , then we call the sequence an diverges to ,denoted bylimn an .(2) If limn infan , then we call the sequence an diverges to ,denoted bylimn an .Theorem Let an be a real sequence. Ifais a Limit point of an , then we havelimn infan a limn useful resultsTheorem Let an be a real sequence, then(1) limn infan limn supan.(2) limn inf an limn supanand limn sup an limn infan(3) If everyan 0, and 0 limn infan limn supan , then wehavelimn sup1an 1limn infanand limn inf1an 1limn Let an and bn be two real sequences.

5 (1) If there is a positive integern0such thatan bn, then we havelimn infan limn infbnand limn supan limn supbn.(2) Suppose that limn infan, limn infbn, limn supan,limn supbn , thenlimn infan limn infbn limn inf an bn limn infan limn supbn(or limn supan limn infbn) limn sup an bn limn supan limn particular, if an converges, we havelimn sup an bn limn an limn supbnandlimn inf an bn limn an limn infbn.(3) Suppose that limn infan, limn infbn, limn supan,limn supbn ,andan 0,bn 0 n, thenlimn infanlimn infbn limn inf anbn limn infanlimn supbn(orlimn infbnlimn supan) limn sup anbn limn supanlimn particular, if an converges, we havelimn sup anbn limn anlimn supbnandlimn inf an bn limn anlimn Let an be apositivereal sequence, thenlimn infan 1an limn inf an 1/n limn sup an 1/n limn supan We can use the inequalities to showlimn n!

6 1/nn 1 Let an be a real sequence, thenlimn infan limn infa1 .. ann limn supa1 .. ann limn Letf: a,d Rbe a continuous function, and an is a real sequence. Iffisincreasing and for everyn, limn infan, limn supan a,d , thenlimn supf an flimn supanand limn inff an flimn :(1) The condition thatfis increasing cannot be removed. Forexample,f x |x|,andak 1/kifkis even 1 1/kifkis odd.(2) The proof is easy if we list the definition of Limit sup and Limit inf. So, weomit Let an be a real sequence satisfyingan p an apfor alln,p. Show that ann :Consider its Limit inf.