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Solution to Understanding Analysis Exercise 2.3


Exercise 2.3.1

(a) Let ε>0 be arbitrary. Since (xn)0, there exists NN such that for all n>N we have xn=|xn|<ε2.Hence xn<ε. Therefore, for all n>N, we also have|xn0|=xn<ε.Thus we have (xn)0.

(b) If x=0 by part (a) we are done. If x0, since xn0 for all n, by Order Limit Theorem, we have x0. Since x0, we have x>0.

Let ε>0 be arbitrary. Since (xn)x, there exists NN such that for all n>N we have |xnx|<εx. Therefore, for all n>N, we have|xnx|= |xnx|xn+x |xnx|x< εxx=ε.


Exercise 2.3.2

(a) Let ε>0 be arbitrary. Since (xn)2, there exists NN such that for all n>N we have |xn2|<32ε.Therefore, for all n>N, we have |2xn131|= |2(xn2)3|= 23|xn2|< 2332ε=ε.Therefore (2xn13)1.

(b) Let 1>ε>0 be arbitrary. Since (xn)2, there exists NN such that for all n>N we have |xn2|<ε.Since ε<1, we get xn>2ε>1.Therefore, for all n>N, we have |1xn12|= |(xn2)2xn|= 12|xn2|< 12ε<ε.Here we used that xn>1. Therefore (1xn)12.


Exercise 2.3.3

Let ε>0 be arbitrary.
Since (xn)l, there exists NN such that for all n>N we have
|xnl|<ε.Hence for all n>N we have(1)ε<xnl<εlε<xn<l+ε. Similarly, since (zn)l, there exists NN such that for all n>N we have
|znl|<ε.Hence for all n>N we have(2)ε<znl<εlε<zn<l+ε.
Note that xnynzn, therefore by (1) we have (3)lε<xnynfor all n>N.
By (2) we have (4)ynzn<l+ε for all n>N.
Combining (3) and (4), we havelε<yn<l+ε|ynl|<εall n>N. Therefore (yn)l as well.


Exercise 2.3.4

Note that (an)0.

(a) By Algebraic Limit Theorem, we have lim(1+2an)=1+2liman=1, lim(1+3an4an2)=1+3liman4limanliman=1. Therefore using Algebraic Limit Theorem again, lim(1+2an1+3an4an2)= lim(1+2an)lim(1+3an4an2)= 11=1.
(b) By Algebraic Limit Theorem, we havelim(an+2)24an= lim(an2+4an+4)4an= liman2+4anan= liman(an+4)an= lim(an+4)=4.
(c) By Algebraic Limit Theorem and multiplying both numerator and denominator by an, we havelim2/an+31/an+5= lim2+3an1+5an= lim(2+3an)lim(1+5an)= 21=2.


Exercise 2.3.5

We first show “if” part. Set limxn=limyn=a. Let ε>0 be arbitrary. Since limxn=a, there exists N1N such that for all n>N we have (5)|xna|<ε.Similarly, since limyn=a, there exists N2N such that for all n>N we have (6)|yna|<ε.Let N=2max{N1,N2}+2, we show that|zna|<εfor all n>N. If n=2k1 is odd, then k>N1. By (5) we have |zna|=|xka|<ε.If n=2k is even, then k>N2. By (6) we have |zna|=|yka|<ε.Therefore (zn) is convergent to the same limit of (xn).

We then “show only if” part. Suppose limzn=a. Let ε>0 be arbitrary. There exists NN such that (7)|zna|<ε for all n>N. Note that for all n>N we also have 2n1>N and 2n>N. Hence by (7) we have|xna|=|z2n1a|<εand |yna|=|z2n1a|<ε.Therefore limxn=limyn=a.


Exercise 2.3.6

We are going to using the following useful formula,ab=aba+b.
By this formula, we can rewrite bn as follows,bn= nn2+2n= n2(n2+2n)n+n2+2n= 2nn+n2+2n=21+n2+2n/n= 21+1+2n.Then since lim2n=2lim1n=20=0, and by Exercise 2.3.1, we havelim1+2n=1.Hence by the Algebraic Limit Theorem, we havelimbn= lim21+1+2n= 21+lim1+2n= 21+1=1.


Exercise 2.3.7

(a) Here is an example. Let (xn) be the sequence (1,0,1,0,1,0,……). Let (yn) be the sequence (-1,0,-1,0,-1,0,……). Then both (xn) and (yn) are divergent. However, the sequence (xn+yn) is given by (0,0,0,0,0,). Therefore (xn+yn) is convergent.

(b) This is impossible. Suppose (xn+yn) and xn are convergent. Note that yn=(xn+yn)xn, therefore by Algebraic Limit Theorem, we have (yn) is convergent.

(c) Let bn=(1)nn, then it is clear that (bn)0. However, we have1bn=(1)n.Since (bn) is not bounded, therefore by Theorem 2.3.2, bn is divergent.

(d) This is impossible. Suppose anbn is bounded, namely there exists M1N such that for all nN, |anbn|M1. Since (bn) is convergent, by Theorem 2.3.2, (bn) is bounded, namely there exists M2N such that for all nN, |bn|M2.Hence for all nN, we have|an||anbn|+|bn|M1+M2.Thus (an) is bounded.

(e) Take (an) to be the sequence (0,0,0,0,……). Take (bn) to be the sequence (1,2,3,4,5,……). Clearly, (an) and (anbn) are convergent but (bn) is not since (bn) is not even bounded.


Exercise 2.3.8

(a) By the Algebraic Limit Theorem, we can show by induction that if (xn)x, then (xnk)xk for any kN. Let p(x) be a polynomial,p(x)=amxm+am1xm1++a1x+a0, where am are real numbers. Then by the Algebraic Limit Theoremlimp(xn)= lim(amxnm+am1xnm1++a1xn+a0)= limamxnm+limam1xnm1++lima1xn+lima0= amlimxnm+am1limxnm1++a1limxn+a0= amxm+am1xm1++a1x+a0=p(x).(b) Consider the function f:RR such that f(0)=1 and f(x)=0,x(,0)(0,).Then take x=0, it is clear that if (xn)0, then (f(xn))0f(0)=1.


Exercise 2.3.9

(a) Let ε>0 be arbitrary. Since (an) is bounded, there exists some M>0 such that (8)|an|<M for all nN.
Since limbn=0, there exists NN such that for all n>N we have(9)|bn|<εM.Therefore it follows from (8) and (9) that for all n>N |anbn0|= |anbn|M|bn|< MεM=ε.We cannot use Algebraic Limit Theorem as the limit of (an) may not exist.

(b) Note that an=anbnbn,anbn=anbn hence it follows from parts (iii) and (iv) of Algebraic Limit Theorem that (an) is convergent if and only if (anbn) is convergent. Hence limbn0 is very important.

(c) Since liman=0 and limbn exists, therefore (bn) is bounded. Then by part (a) we have that limanbn=0=ab as a=0. Note that in order to use part (a), the sequences (an) and (bn) here correspond to the sequence (bn) and (an) in part (a), respectively.


Exercise 2.3.10

(a) Counterexample. Let (an) and (bn) be the same sequence (1,2,3,4,5,). Then (anbn) is the sequence (0,0,0,). It is clear that lim(anbn)=0. However the limits of (an) and (bn) do not exist since they are not even bounded.

(b) We use the following inequality ||x||y|||xy| to prove it.
Let ε>0 be arbitrary. Since (bn)b, there exists NN such that for all n>N we have |bnb|<ε. Then by our inequality ||x||y|||xy|, for all n>N we have||bn||b|||bnb|<ε.Hence (|bn|)|b|.

(c) Yes. Note thatbn=(bnan)+an,hence by Algebraic Limit Theorem we havelimbn=lim(bnan)+liman=0+a=a.
(d) Yes. Note that |bnb|an implies an0 and anbnban.Since liman=0, by Algebraic Limit Theorem we have lim(an)=0. Now by the Sequeeze Theorem, we have lim(bnb)=0. Hence limbn=lim(bnb)+limb=0+b=b.


Exercise 2.3.11

(a) Let ε>0 be arbitrary. Since (xn) is convergent, by Theorem 2.3.2, (xn) is bounded. Hence there exists M>0 such that |xnx|<M for all nN. Again since (xn)x, there exists N0N such that for all n>N0 we have |xnx|<ε2. Now we take an integer N>2N0Mε, we are going to show that for all n>N, we have |ynx|<ε. Indeed, for n>N, we have|x1x|++|xN0x|<N0M. |xN0+1x|++|xnx|<(nN0)ε2. Therefore (10)|x1x|++|xnx|<N0M+(nN0)ε2 Since n>N>2N0Mε, we have (11)N0M<nε2. Therefore|ynx|=|x1++xnnx|n |x1x|++|xnx|nuse (10)< N0M+(nN0)ε2nuse (11)< nε2+(nN0)ε2n<ε.Hence (yn)x as well.

(b) Take xn=(1)n, then (xn) does not converge but (yn)0.


Exercise 2.3.12

(a) True. Take any element bB. Since every an is an upper bound of B, we have anb for all nN. Therefore by the Order Limit Theorem, we have a=limanb.Hence ab. Since b is chosen arbitrarily, we see that a is an upper bound of B.

(b) True. Suppose a(0,1), then take ε=min{a,1a}, then aε>0 and a+ε<1. Consider the ε-neighborhood Vε(a), thenVε(a)(0,1). Recall the definition 2.2.3B, if (an)a, then there exists NN such that anVε(a) for all n>N. This implies an(0,1) for all n>N which contradicts our assumption. Hence our assumption that a(0,1) is impossible. Thus a is in the complement of (0,1).

(c) False. We are going to use Theorem 1.4.3 to construct a counter example. Let a=2 which is irrational. By Theorem 1.4.3, for any nN we can find a rational number an such that2<an<2+1n.Then we show that (an)2.

Let ε>0 be arbitrary. Let N>1ε, then 1N<ε. For all n>N, we have |an2|<1n<1N=ε.Hence (an)a. By our choice all an are rational, but the limit is 2 which is irrational.


Exercise 2.3.13

(a) Note that amn=11+nm, hence limmamn=1 andlimn(limmamn)=limn1=1.Similarly, amn=mn1+mn, hence limnamn=0 and limm(limnamn)=limm0=0.
(b) Yes. It is cleat that limm,namn=0. Similarly to part (a), we havelimm(limnamn)=limn(limmamn)=0.The three limits are the same.
If amn=mnm2+n2, thenlimm(limnamn)=limn(limmamn)=0.However limm,namn does not exist. Suppose the limit exists. Taking m=n>N, we have ann=12. Hence this sequences has infinitely many one-half’s. By the negation of Exercise 2.2.4(b), the limit has to be 12. Taking m=2n>2N, then we have a2n,n=25. Hence this sequences has infinitely many two-fifth’s. By the negation of Exercise 2.2.4(b), the limit has to be 25. But this limit should be unique, hence we get a contradiction and we are done.

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