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MATH 409, Summer 2019,

Practice Problem Set 2

F. Baudier (Texas A&M University)

May 29, 2019

1 Warm up

Exercise1.Show that a sequence (xn)1n=1is convergent to`2R, if and only if for every" >0 there existsN2Nsuch that for alln>N,jxn`j6".

Hint.Exploit the denition.Possible solution.

Exercise2.Show that a sequence (xn)1n=1is convergent to`2R, if and only if for every"2(0;2) there existsN2Nsuch that for alln>N,jxn`j< ".

Hint.Exploit the denition.Possible solution.

Exercise3.Show that a sequence (xn)1n=1is convergent to`2R, if and only if for every" >0 there existsN2Nsuch that for alln>N,jxn`j<256".

Hint.Exploit the denition.Possible solution.

Exercise4.Let (xn)1n=1and (yn)1n=1be convergent sequences. Show that 1. the sequence ( xn+yn)1n=1is convergent and that lim n!1(xn+yn) = limn!1xn+ limn!1yn: 1

2.the sequence (2 xn5yn)1n=1is convergent and that

lim n!1(2xn5yn) = 2 limn!1xn5 limn!1yn: 3. the sequence ( xnyn)1n=1is convergent and that lim n!1(xnyn) = limn!1xnlimn!1yn: Hint.Use the denition of convergence and the algebraic equality,abcd= b(ac) +c(bd) (for 3)Possible solution.Assume that limn!1xn=`1<1and limn!1yn=`2< 1. 1. Let " >0, then there existN1;N22Nsuch that for allnN1,jxn`1j< "2 and for allnN2,jyn`2j<"2 . If follows from the triangle inequality thatjxn+yn(`1+`2)j=jxn`1+yn`2j jxn`1j+jyn`2j, and hence fornmaxfN1;N2g,jxn+yn(`1+`2)"2 +"2 2. If follo wsfrom the triangle inequalit ythat jxnyn(`1`2)j=j(xn`1)yn+

1(yn`2)j jxn`1jjynj+jyn`2jj`1j. Since (yn)1n=1is convergent,

and thus bounded, there existsM >0 such that for alln2N,jynj M. Let" >0. Ifj`1j>0, then there existN1;N22Nsuch that for all nN1,jxn`1j<"2Mand for allnN2,jyn`2j<"2j`1j, and hence fornmaxfN1;N2g,jxnyn(`1`2)j<"2MM+"2j`1jj`1j=". If j`1j= 0 then fornmaxfN1;N2g,jxnynj<"2MM < ", and the proof is complete.Exercise5.Show that (13 n)1n=1converges and compute limn!113 n.

Hint.Try to use the idea of the proof of 3. in Example 1.Possible solution.It follows from the Archimedean Principle that for every" >0

there existsN2Nsuch that 0<1" < N. It can be easily shown by induction (do it!) thatn3nfor alln2N. FornN,j13 n0j=13 n1n 1N < "and lim n!113 n= 0.2

2 Useful results about sequences

Exercise6.Let (xn)1n=1be a sequence of real numbers and`2R. Show that lim n!1xn=`if and only if limn!1jxn`j= 0. Hint.Simply consider the sequences (xn)1n=1and (yn)1n=1= (jxn`j)1n=1, and

apply the denition of convergence.Possible solution.Assume that limn!1xn=`. Let" >0, then there exists

N2Nsuch that for allnN,jxn`j< ". Butjxn`j 0=jxn`jand fornN,jxn`j 0< ", and thus limn!1jxn`j= 0.

Assume now that lim

n!1jxn`j= 0, Let" >0, then there existsN2N such that for allnN,jxn`j 0< "butjxn`j 0=jxn`jand for

nN,jxn`j< ". Therefore, limn!1xn=`.Exercise7.Let (xn)1n=1and (yn)1n=1be two sequences of real numbers and

`2R. Assume that limn!1xn= 0 and that there existsN2Nso that for all n>Nwe havejyn`j6jxnj. Show that limn!1yn=`.

Hint.Exploit the denition of convergence.Possible solution.Assume that limn!1xn= 0 and that there existsN2Nso

that for alln>Nwe havejyn`j6jxnj. Let" >0. Then thereN12Nsuch that for allnN1,jxnj< ". IfnmaxfN;N1gthen,jyn`j jxnj< ".

Therefore, lim

n!1yn=`.Exercise8.Let (xn)1n=1and (yn)1n=1be sequences of real numbers. Assume that (xn)1n=1is bounded and that limn!1yn= 0. Let (zn)1n=1= (xnyn)1n=1.

Show that lim

n!1zn= 0 Hint.Exploit the denitions of convergence, boundedness, and the properties of the absolute value.Possible solution.Assume that there existsM0 such that for alln2N, jxnj Mand that limn!1yn= 0. IfM= 0jxnynj= 0 and the conclusion clearly holds. Otherwise, if" >0, then there existsN2Nsuch that for all nN,jynj<"M . This yields thatjxnynj=jxnjjynj Mjynj< M"M

whenevernN. Therefore, (zn)1n=1converges to 0.Exercise9.Let (xn)1n=1be a sequence of real numbers. Show that (xn)1n=1is

increasing if and only if for alln2N,xnxn+1. 3 Hint.One implication follows directly from the denition. The other one can be proven using an induction.Possible solution.Assume that (xn)1n=1is increasing, i.e. for allkm,xk x m. Letn2N, then by simply takingk=nandm=n+1, one hasxnxn+1. Assume now that for alln2N,xnxn+1. Since whenk < mone can always writem=k+rfor somer2N, the conclusion will follow if one can prove that for allk;r2N,xkxk+r. Letk2Nand forr2NletP(r) be the statement:xkxk+r. Our assumption says thatP(1) is true. Assume now thatP(r) is true. On one hand,xkxk+rby our induction hypothesis. On the other hand,xk+rxk+r+1by our assumption, and hence by transitivity of the order relationxkxk+r+1andP(r+ 1) is true. By the Principle of Mathematical InductionP(r) is true for allr2N. Sincekwas xed but arbitrary, one just proved that for allk;r2N,xkxk+rand the conclusion follows.3 Around the Monotone Convergence Theorem Exercise10.Let (xn)1n=1be a sequence of real numbers. Show without using the Monotone Convergence Theorem that if (xn)1n=1is decreasing and bounded below then (xn)1n=1is convergent. Hint.You could mimic the proof of the increasing version and use the ap- proximation property of inma to show that (xn)1n=1converges to inffxn:n2

Ng.Possible solution.

Exercise11.Show that ifjaj<1 then limn!1an= 0.

Hint.Use the Monotone Convergence Theorem.Possible solution.Assume thatjaj<1, thenjajn+1=jajna (jajn)1n=1is strictly decreasing. It is clear that (jajn)1n=1is bounded below by 0, and by the Monotone Convergence Theorem, (jajn)1n=1is convergent. Denote` the limit. Then lim n!1jajn= limn!1jajn+1=`and sincejajn+1=jajna by the basic manipulations of limits`satises the equation`=`jaj. The only solutions are`= 0 orjaj= 1 and the second alternative is impossible, thus lim n!1jajn= 0. We conclude with the Squeeze Theorem since for alln2N, jajnan jajn.4 Exercise12.Let (xn)1n=1be a bounded sequence of real numbers. For alln2N, lettn:= inffxk:kng. Show that (tn)1n=1is convergent. Hint.You could use the Monotone Convergent Theorem and mimic the proof of Lemma 7 in the lecture notes.Possible solution.Letn2N. Sincefxk:kng fxk:kn+ 1g,tn= inffxk:kng inffxk:kn+ 1g=tn+1, and (tn)1n=1is increasing. Since (xn)1n=1is bounded, (tn)1n=1is also bounded. By the Monotone Convergence Theorem (tn)1n=1is convergent.Exercise13 (The Nested Interval Theorem).Recall that a sequence of set (An)n2Nis nested if for alln2N,An+1An. Recall also that a closed interval is a subset ofRof the form [a;b]. Show that a nested sequence of closed intervals has a non-empty intersection. Hint.You could use the Least Upper Bound Theorem or the Monotone Con- vergent Theorem.Possible solution.

Exercise14.Let 0< x1< y1and set for alln2N,

x n+1=px nynandyn+1=xn+yn2 i)

Pro vethat for all n2N, 0< xn< yn.

ii) Pro vethat ( xn)1n=1is increasing and bounded above. iii) Pro vethat ( yn)1n=1is decreasing and bounded below. iv)

Pro vethat for all n2N, 0< yn+1xn+1 n. v)

Pro vethat lim

n!1xn= limn!1yn. This common limit:= limn!1xn= limn!1ynin v) above is called the arithmetic-geometric mean ofx1andy1and has many applications. Hint.For i) use induction. For ii)-iii) use i). For iv) use induction. For v) use the Monotone Convergence Theorem and the Squeeze Theorem.5 Possible solution.(i)F orn= 1 the inequality holds by assumption. Ifn2, then x n=px n1yn1andyn=xn1+yn12 and one needs to show that the geometric mean is smaller that the arith- metic mean i.e.px n1yn1

4xn1yn1. But,

(xn1+yn1)24xn1yn1=x2n1+ 2xn1yn1+y2n14xn1yn1 =x2n12xn1yn1+y2n1= (xn1yn1)20: The conclusion follows since one can easily prove by induction (and we omit the details) that for alln2N,xn;yn>0 andxn6=yn. (ii)

By (i) for all n2N,xn+1=px

nyn>px nxn=jxnj=xnand (xn)1n=1is strictly increasing. (iii)

Similarly for all n2N,yn+1=xn+yn2

Sincex1< y1,px

1y1x1>0 and this yields thaty2x2

Assume now thatyn+1xn+1 n. We need to show thaty1x12 n+1 y n+2xn+2>0. But, y 1x12 n+1yn+2+xn+2=12 y 1x12 nyn+1+xn+12 +px n+1yn+1 12 (yn+1xn+1)yn+1+xn+12 +px n+1yn+1 =px n+1yn+1xn+1>0; and the induction is complete. (v) By (ii), (iii) and the Monotone Con vergenceTheorem b oth( xn)1n=1and (yn)1n=1are convergent. By (iv) and the Squeeze Theorem limn!1(yn x n) = 0 and thus limn!1xn= limn!1yn.4 Subsequences Exercise15.Let (xn)1n=1be a sequence of real numbers anda2R. If the sequence (xn)1n=1does not converge toa, prove that there exists an"0>0 and a subsequence (xnk)1k=1of (xn)1n=1, so thatjxnkaj>"0for allk2N. 6 Hint.Negate the denition of convergence and construct the subsequence re- cursively.Possible solution.Assume that (xn)1n=1does not converge toa. Then, there exists"0>0 so that for everyk2N, there existsnk>kwithjxnkaj>"0:() We shall now construct the subsequence recursively. In particular, fork= 1, there existsn12Nwithjxn1aj>"0. Assume now that there existn1< < nkand (xni)ki=1withjxniaj>"0for 16i6k. By (), there exists N>nk+ 1 withjxNaj>"0. Denenk+1=N. Then,nk+1>nk+ 1> nk,

jxnk+1aj>"0and the recursive construction is complete.Exercise16.Let (xn)1n=1be a sequence of real numbers and`2R. Assume that

for every subsequence (yn)1n=1of (xn)1n=1, there exists a further subsequence (zn)1n=1of (yn)1n=1that converges to`. Prove that the original sequence (xn)1n=1converges to`.

Hint.Argue by contradiction using Exercise 15.Possible solution.If (xn)1n=1does not converge to`, then by Exercise 15 there

exist"0>0 and a subsequence (xnk)1k=1of (xn)1n=1so that jxnk`j>"0for allk2N:() By assumption, (xnk)1k=1has a further subsequence (xnkm)1m=1that converges to`and therefore there ism02Nso that for allm>m0,jxnkm`j< "0. As k m0, for instance, is still inN, by (),jxnkm0`j>"0. This contradiction

completes the proof.Exercise17.For this exercise we will dene a top point of a sequence (xn)1n=1as follows: we say thatxpis a top point of the sequence if for allnp,xnxp.

Prove the monotone subsequence lemma using the notion of top point. Hint.Consider the following three cases: the sequence has innitely many top

points, or nitely many top points, or no top points.Possible solution.Assume rst that (xn)1n=1has no top points. Letk1= 1.

Sincexk1is not a top point there existsk2> k1such thatxk2> xk1. But x k2is not a top point either and there existsk3> k2> k1such thatxk3> x k2> xk1. If we continue this process indenitely we can construct recursively a subsequence (xk1)1n=1that is strictly increasing. Now, assume that a sequence (xn)1n=1has innitely many top points then there existp1< p2<< pk< 7 such that for allmn,xpmxpnand the subsequence (xpk)1k=1is decreasing. If (xn)1n=1has nitely many top points and letxpthe largest of those top points. Letk1=p+ 1, thenxk1is not a top point and hence there existsk2> k1such thatxk2> xk1. Sincexk2is not a top point there existsk3> k2> k1such that x k3> xk2> xk1, and we can construct recursively a subsequence (xkn)1n=1that is (strictly) increasing. In all three cases, we were able to show the existence of a monotone subsequence.Exercise18.Let (xn)1n=1be a bounded sequence of real numbers. Lett:= liminf n!1xn. Show that there exists a subsequence (yn)1n=1of (xn)1n=1such that lim n!1yn=t. Hint.Construct the subsequence recursively using the approximation property for suprema and conclude with the Squeeze Theorem.5 Constructing sequences Exercise19.Prove that for every real numberxthere exists a sequence of rational numbers (qn)1n=1with limn!1qn=x.

Hint.Use the density ofQinRand the Squeeze Theorem.Possible solution.Letxbe a real number. Then by density ofQinR, for every

n2Nthere existsqn2Qsuch thatx < qn< x+1n . By the Squeeze Theorem lim n!1qn=x.Exercise20.LetXbe a non-empty subset ofRthat is bounded above. Assume

that sup(X)=2X. Prove that there exists a strictly increasing sequence (xn)1n=1ofXso that limn!1xn= sup(X).

Hint.Construct the sequence recursively using the approximation property for suprema.Possible solution.Sets= sup(X). We will recursively choose for eachn2Na numberxn2X, so thatx1<< xnandjxnsj<1n . This will yield the desired sequence. To do this rigorously we will use the Principle of Mathematical Induction. LetP(n) be the statement: there existx1<< xnelements inX such thats1n < xn6s. By the approximation property for suprema (for"= 1), we may choose x

12Xwiths1< x16sandP(1) is true.

Assume now that there existx1<< xnelements inXsuch thats1n x n6s. Sincexn6sands =2X, we havexn< s, i.e.sxn>0. Set 8 "= minf1n+1;sxng, which is positive. By the approximation property of suprema we may choosexn+12Xwiths" < xn+16s. Sinces" < xn+16 s < s+", we concludejxn+1sj< "61=(n+ 1). Furthermore, observe that x n=s(sxn)6s" < xn+1, thereforexn+1satises the desired properties. By the Principle of Mathematical Induction for alln2NP(n) is true, i.e. for everyn2Nthere existx1<< xnelements inXsuch thats1n < xn6s. The sequence (xn)1n=1is the desired sequence, since by the Squeeze Theorem lim n!1xn=s.Exercise21.LetXbe a non-empty subset ofRthat is bounded below. Assume

that inf(X)=2X. Prove that there exists a strictly decreasing sequence (xn)1n=1ofXso that limn!1xn= inf(X).

Hint.Mimic the proof of Exercise 20.6 Cauchy Sequences Exercise22.Show that a Cauchy sequence is bounded. Hint.The proof is similar to the proof of the fact that a convergent sequence is bounded.Possible solution.Assume that (xn)1n=1is Cauchy. Then for"= 1 there exists N2Nsuch that for alln;mN,jxnxmj 1. In particular form=N and by reverse triangle inequalityjxnj 1 +jxNjfor allnN. LetM:=

maxfjx1j;jx2j;:::;jxN1j;1 +jxNjg, then for alln2N,jxnj Mand (xn)1n=1is bounded.Exercise23.Let (xn)1n=1and (yn)1n=1be two Cauchy sequences such thatjynj

>0 for alln2N. Show that the sequence (xny n)1n=1is Cauchy.

Hint.Use the Triangle Inequality and ad-hoc algebraic manipulations.Possible solution.If follows from the triangle inequality and the assumptions

thatjxny nxmy mj=jxnymynxmy nymj=j(xnxm)ym(ynym)xmy nymj jxnxmjjynj 2+ jynymjjxmj

2. Since a Cauchy sequence is bounded (cf Exercise 22) there exists

M >0 such that for alln2N, maxfjynj;jxnjg M. Let" >0. Then there existN1;N22Nsuch that for alln;mN1,jxnxmj<"22Mand for all n;mN2,jynymj<"22M, and hence forn;mmaxfN1;N2g,jxny nxmy mj jxnxmjjynj

2+jynymjjxmj

22"22M+M

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