MATH 4310 :: Introduction to Real Analysis I :: Spring ...

1 MATH 4310 :: Introduction to real Analysis I :: Spring 2015 :: LangouHomework #5 KeyHW5 , , , (first try due on Tuesday March 3rd, second try due on Tuesday March 10th)In this homework, we are moving one notch further in our mastery of Analysis . While youare expected to be able to answer all of these questions starting from first principle (=definitions), most of these questions are much better handled with some of our theorems,corollary, etc. Exercise is a good example. Note: Soon, the only way to solve problemswill be to use powerful theorems and (sn)n Nbe the sequence of number in Fig. listed in the indicated the setSof subsequential limits of (sn)n {1n,for alln N} {0}.Note: we know that 0 is a subsequential limit of (sn)n N(that is to say we know that 0 is inS ) byusing the definition of subsequential limit.

2 0 is a subsequential limit of (sn)n Nbecause we can finda subsequence of (sn)n Nthat converges to 0. However there is another way to know that 0 is states that a limit of a sequence of subsequential limits is a subsequential limit :Theorem : LetSdenote the set of subsequential limits of a sequence (sn)n N. Suppose(tn)n Nis a sequence inS Rand thatt= limtn. Thentbelongs our case, (i) for alln N,1n S, (ii) lim1n= 0, so, with (i) and (ii) and Theorem , we get that0 lim supsnand lim infsn= 0 and lim supsn= 1 Note:There can be two ways to go about finding lim supsn. Either you use the definition of lim page 60 Equation (1).lim supsn= limn ( Sup{smfor allm > n}).In which case, you find lim supsn= you use Theorem page 74:Theorem : Let (sn)n Nbe a sequence.

3 LetSdenote the set of subsequential limitsof (sn)n N. Thenlim supsn= SupS.(Note that computingSis asked in part ) In which case, you find lim supsn= ideato find twice the same value .. (Same comment for lim inf.) that lim sup|sn|= 0 if and only limsn= 0 .Since we know that, (Exercise ,)lim|sn|= 0 if and only limsn= 0,we will prove thatlim sup|sn|= 0 if and only lim|sn|= :we prove the fact that:lim|sn|= 0 if and only limsn= 0 >0, N, n > N,|sn 0|< >0, N, n > N,||sn| 0|< lim|sn|= 0(1) By Theorem , (applied to|sn|,) we have thatlim|sn|= 0 lim sup|sn|= : If limsnis defined [as a real number, + , or ], then lim infsn=limsn= lim supsn. We want to prove thatlim sup|sn|= 0 lim|sn|= us assume that lim sup|sn|= 0.

4 (a) We have that, for alln, 0 |sn|. This leads to 0 lim inf|sn|.Note:we prove the fact that:Let ( n N, sn) then( lim infsn).Let R. n N, sn, N N, nsuch thatn > N, sn, N N, is a lower bound for{snsuch thatn > N}, N N, Inf{snsuch thatn > N}, limN ( Inf{snsuch thatn > N}), lim infsn.(b) We also have that lim inf|sn| lim sup|sn|.Note:we prove the fact that:lim inf|sn| lim sup|sn|.We start with the fact that, for any nonempty subsetS, we have InfS SupS, whereSis the set{snsuch thatn > N}for a fixedN. N N,( Inf{snsuch thatn > N} Sup{snsuch thatn > N}), limN ( Inf{snsuch thatn > N}) limN ( Sup{snsuch thatn > N}), lim inf|sn| lim sup|sn|.So, combining0 lim inf|sn|lim inf|sn| lim sup|sn|lim sup|sn|= 0leads us to:0 lim inf|sn| lim sup|sn|= 0,which leads tolim inf|sn|= lim sup|sn|= Theorem , we conclude that lim|sn|= : If lim infsn= lim supsn, then limsnis defined and lim infsn= limsn=lim (sn)n Nis bounded if and only if lim sup|sn|<+.

5 Assume that (sn)n Nis , letMbe a real number such that, for allm,|sm| , for alln, Sup{|sm|such thatm > n} , lim ( Sup{|sm|such thatm > n}) , lim supsn we get lim supsn<+ . Assume that lim sup|sn|<+ . So letMbe a real number such that lim sup|sn|=M. Sincelim ( Sup{|sm|such thatm > n}) =M, there existsN Nsuch that|Sup{|sm|such thatm > N} M|<1,this implies thatSup{|sm|such thatm > N}< M+ 1,so this means that,for allm > N,|sm|< M+ 1,so this means that, for alln,|sn| max (|s1|,..,|sN|,M+ 1).This means that (sn)n Nis way to prove . We can look at the contrapositive. We want to prove that if (sn)n Nis notbounded then lim sup|sn|= + . If (sn)n Nis not bounded, then by Theorem , (sn)n Nhas asubsequence with limit +.

6 Since lim supsnis an upper bound for the set of all subsequential limits of(sn)n N(by Theorem ) and that + belongs toSthen, lim sup|sn|= + . (The fact that + belongs toS is not really rigorous as far as the book defines + , so this proof is a little so so.)Theorem : Let (sn)n Nbe a sequence. If the sequence (sn)n Nis not bounded above,it has a subsequence with limit + .Theorem : Let (sn)n Nbe a sequence. LetSdenote the set of subsequential limitsof (sn)n N. Thenlim supsn= (n!)1 (sn)n Nbe such that, for alln,sn=n!, then our problem of studying the convergence of (n!)1/nbecomes the problem of studying the convergence of (s1/nn)n us have a look atsn+1sn:sn+1sn=(n+ 1)!n!=n+ 1,solimsn+1sn= + .We can then use Corollary , page 81:Corollary : If lim|sn+1sn|exists [and equalsL], then lim|sn|1/nexists [and equalsL].

7 To conclude thatlim(n!)1/n= + . (n!)1 (sn)n Nbe such that, for alln,sn=n!nn, so that our problem of studying the convergence of1n(n!)1/nbecomes the problem of studying the convergence of (s1/nn)n us have a look atsn+1sn:sn+1sn=(n+ 1)!(n+ 1)n+1nnn!=(nn+ 1)n=enlnnn+1=enln(1 1n+1)=en( 1n+1+O(1n2))=e 1+O(1n) n e 1,solimsn+1sn=e can then use Corollary , page 81:Corollary : If lim|sn+1sn|exists [and equalsL], then lim|sn|1/nexists [and equalsL].to conclude thatlim1n(n!)1/n=e 1.