AoPS Community Chapter 3 Selected Exercises (Rudin)

AoPS Community Chapter 3 Selected Exercises (Rudin)W. Rudin, Principles of Mathematical Analysis, 3rd Editionwww.artofproblemsolving.com/community/c441003by Solumilkyu

1. Prove that convergence of {sn} implies convergence of {|sn|}. Is the converse true?—–(i) Suppose {sn} converges to some s, then for ε > 0, there exists an integer N such that|sn − s| < ε for all n ≥ N . It follows that ||sn| − |s|| ≤ |sn − s| < ε for n ≥ N , and then {|sn|}converges to |s|.(ii) No. For example, let sn = (−1)n for n = 1, 2, . . ..

2. Calculate limn→∞

(√n2 + n− n

).—–We have

limn→∞

(√n2 + n− n

)= lim

n→∞

[(√n2 + n− n

)·√n2 + n+ n√n2 + n+ n

]

= limn→∞

(n2 + n− n2√n2 + n+ n

)= lim

n→∞

(n√

n2 + n+ n

)

= limn→∞

1√1 + 1

n + 1

=

1

2

3. If s1 =√

2, andsn+1 =

√2 +√sn (n = 1, 2, 3, . . .)

prove that {sn} converges, and that sn < 2 for n = 1, 2, 3, . . ..—–First, it is easy to prove that sn < 2 for n = 1, 2, . . ., and that {sn} is bounded above. To show it,use induction on n. At first, s1 =

√2 < 2 is immediate. Now suppose sn < 2 for some positiveinteger n, then

sn+1 =√

2 +√sn <

√2 +√

2 < 2

© 2019 AoPS Incorporated 1

AoPS Community Chapter 3 Selected Exercises (Rudin)Hence sn < 2 for all n. Next, we show that {sn} is monotonically increasing, and so the con-vergence of {sn} follows by Theorem 3.14, completing this exercise. Use induction again, atfirst, clearly s1 < s2. Now suppose sn ≤ sn+1 for some positive integer n, then

sn+1 =√

2 +√sn ≤

√2 +√sn+1 = sn+2

Hence {sn} is monotonically increasing.4. Find the upper and lower limits of the sequence {sn} defined by

s1 = 0; s2m =s2m−1

2; s2m+1 =

1

2+ s2m

—–By some observation, we show that, for m = 1, 2, . . .,

s2m =1

2− 1

2mand s2m+1 = 1− 1

2m

Note that the second equality is a direct consequence of the first one, so we only have toprove the first. Use induction on m, for m = 1, we have s2 = s12 = 0 = 1

2 −121

. Now supposes2m = 1

2 −12m for some integer m ≥ 1, then

s2(m+1) = s2m+2

=s2m+1

2

=1

2

(1

2+ s2m

)=

1

2

(1− 1

2m

)=

1

2− 1

2m+1

Hence s2m = 12 −

12m for all m. The above argument gives the following results

lim supn→∞

sn = 1 and lim infn→∞

sn =1

2

5. For any two real sequences {an}, {bn}, prove thatlim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn

provided the sum on the right is not of the form∞−∞.—–


AoPS Community Chapter 3 Selected Exercises (Rudin)W.l.o.g., we may consider the following three cases, where the first two follow by the assump-tion that∞−∞ (and −∞+∞) is exceptional.Case 1: If lim supn→∞ an = +∞ or lim supn→∞ bn = +∞, then the r.h.s. of the inequality mustbe either of the form∞+∞ or∞+ r, for some real number r. Hence the result is trivial.Case 2: If lim supn→∞ an = −∞ and {bn} has an upper bound M . Then for each real number x,there exists N such that n ≥ N implies an < x, and so

an + bn ≤ an +M < x+M

Since x is arbitrary, we have limn→∞ (an + bn) = −∞, i.e., lim supn→∞ (an + bn) = −∞. Hencethe result follows.Case 3: If lim supn→∞ an = a and lim supn→∞ bn = b, where a, b ∈ R. Suppose lim supn→∞ (an + bn) >a + b, then we can choose a subsequence {nk} of increasing positive integers such thatlimk→∞ (ank

+ bnk) > a+b. Choose a numberx0 such that limk→∞ (ank

+ bnk) > x0 > a+b, thenthere existsK such that k ≥ K implies ank

+bnk> x0. On the other hand, since x0

2 + a−b2 > a and

x02 + b−a

2 > b, by Theorem 3.17(b), there existK1 andK2 such that k ≥ K1 implies ank< x0

2 +a−b2 ,and k ≥ K2 implies bnk

< x02 + b−a

2 . Put K ′ = max{K1,K2}, then k ≥ K ′ impliesank

+ bnk<

(x02

+a− b

2

)+

(x02

+b− a

2

)= x0

Put M = max{K,K ′}, then k ≥ M implies both ank+ bnk

> x0 and ank+ bnk

< x0. This givesa contradiction.6. Investigate the behavior (convergence or divergence) of∑ an if(a) an =

√n+ 1−

√n;

(b) an =√n+1−

√n

n ;(c) an = ( n√n− 1)

n;(d) an = 1zn , for complex values of z.—–

(a) Sincem∑n=1

an =(√

2−√

1)

+(√

3−√

2)

+ · · ·+(√m+ 1−

√m)

=√m+ 1− 1

we have∑ an = limm→∞(√m+ 1− 1

)=∞, i.e.,∑ an diverges.—–(b) Since

√n+ 1−

√n

n=

1

n(√n+ 1 +

√n) ≤ 1

2n√n

=1

2n32


AoPS Community Chapter 3 Selected Exercises (Rudin)The series∑ an converges by the comparison test with the p-series∑ 1

2n32

, where p = 32 .—–(c) Since

limn→∞

n√an = lim

n→∞

(n√n− 1

)= 1− 1 = 0

where the third equality follows by Theorem 3.20(c). By the root rest,∑ an converges.—–(d) We skip this question.7. Prove that the convergence of∑ an implies the convergence of∑ √

ann

if an ≥ 0.—–If an ≥ 0 for n = 1, 2, . . ., then by arithmetic-geometry inequality, we have

√ann≤ 1

2

(an +

1

n2

)So if∑ an converges, then adding the convergence of∑ 1

n2 , and by Theorem 3.47, that∑ 1

2

(an +

1

n2

)=

1

2

(∑an +

∑ 1

n2

)the series∑ 1

2

(an + 1

n2

) also converges. Hence∑ √ann converges by the comparison test.

8. If∑ an converges, and if bn is monotonic and bounded, prove that∑ anbn converges.—–Since {bn} is monotonic and bounded, we may only consider the following two cases.Case 1: If bn ↘ b, put cn = bn − b. Then cn ↘ 0, which follows by Theorem 3.42 that ∑ ancnconverges. Also, by Theorem 3.47,∑ ban converges. Finally since we can write∑

anbn =∑

an (cn + b) =∑

(ancn + ban)

By Theorem 3.47 again,∑ anbn converges.Case 2: If bn ↗ b, put dn = b− bn. Then the argument is similar to that in Case 1.10. Suppose that the coefficient of the power series∑ anz

n are integers, infinitely many of whichare distinct from zero. Prove that the radius of convergence is at most 1.—–


AoPS Community Chapter 3 Selected Exercises (Rudin)Since the sequence {an} contains infinitely many nonzero numbers, we may let {nk} be asubsequence of positive integers such that ank

6= 0 for each k. Moreover, since an is an integerfor each n, we have |ank| ≥ 1 for each k, and thus nk

√|ank| ≥ 1. Hence lim supn→∞

n√|an| ≥ 1or

R =1

lim supn→∞

n√|an|≤ 1

11. Suppose an > 0, sn = a1 + · · ·+ an, and∑ an diverges.(a) Prove that∑ an1+an

diverges.(b) Prove thataN+1

sN+1+ · · ·+ aN+k

sN+k≥ 1− sN

sN+k

and deduce that ansn

diverges.(c) Prove thatans2n≤ 1

sn−1− 1

sn

and deduce that ans2n

converges.(d) What can be said about ∑ an

1 + nanand ∑ an

1 + n2an?

—–(a) To show this, we only have to consider the following two cases.Case 1: If {an} is not bounded above, then limn→∞

an1+an

6= 0. Hence∑ an1+an

diverges.Case 2: If {an} is bounded above, choose a fixed number M > 0 such that an ≤ M for alln. Then an

1+an≥ an

1+M for all n. Since ∑ an1+M apparently diverges, the divergence of ∑ an

1+anfollows by the comparison test.—–(b) Since an > 0 for each n, the sequence {sn} is monotonically increasing. ThusaN+1

sN+1+aN+2

sN+2+ · · ·+ aN+k

sN+k≥ aN+1 + aN+2 + · · ·+ aN+k

sN+k

=sN+k − sNsN+k

= 1− sNsN+k

Now, to show that ∑ ansn

diverges, let tn = a1s1

+ a2s2

+ · · · + ansn

. Since {sn} is monotonicallyincreasing, and since ∑ an diverges, {sn} is not bounded above. So given a positive integer


AoPS Community Chapter 3 Selected Exercises (Rudin)N , there is an another positive integer M with M > N , such that n ≥ M implies sn > 2sN , orsNsn< 1

2 . It follows thattM − tN =

aN+1

sN+1+aN+2

sN+2+ · · ·+ aM

sM≥ 1− sN

sM> 1− 1

2=

1

2

and then the sequence {tn} is not Cauchy. So {tn} diverges, i.e.,∑ ansn

diverges.—–(c) First, the inequality follows by writingans2n

=sn − sn−1

s2n≤ sn − sn−1

sn−1sn=

1

sn−1− 1

sn

Now, to show that∑ ans2n

converges, observe thatn∑k=1

aks2k

=a1a21

+

n∑k=2

aks2k≤ 1

a1+

n∑k=2

(1

sk−1− 1

sk

)=

2

a1− 1

sn

Since sn ↗∞, the r.h.s. of the above inequality converges to 2a1

as n→∞ (this is a telescop-ing series (https: // en. wikipedia. org/ wiki/ Telescoping_ series )). So we conclude that∑ an

s2nconverges.—–(d) (i) The series ∑ an

1+nanmay be either convergent or divergent, we shall give two examplesfor them.Example 1: Let an = 1

n for n = 1, 2, . . .. Then∑ an diverges, and∑ an1+nan

=∑ 1

2n also diverges.Example 2: Let

an =

1√n

if n is a perfect square;1

n2if otherwise

Note that∑ an still diverges. However,an

1 + nan≤

1

nif n = 1, 4, 9, 16, 25, . . .;

1

n2if otherwise

which follows that∑ an1+nan

converges.(ii) The series∑ an

1+n2anapparently converges, since an

1+n2an≤ an

n2an= 1

n2 .12. Suppose an > 0 and∑ an converges. Put

rn =

∞∑m=n

am


AoPS Community Chapter 3 Selected Exercises (Rudin)(a) Prove that

amrm

+ · · ·+ anrn≥ 1− rn

rm

if m < n, and deduce that∑ anrn

diverges.(b) Prove thatan√rn

< 2 (√rn −

√rn+1)

and deduce that∑ an√rn

converges.—–(a) First, the inequality follows by writing, for m < n,

amrm

+am+1

rm+1+ · · ·+ an

rn≥ am + am+1 + · · ·+ an

rm

>am + am+1 + · · ·+ an−1

rm

=rm − rnrm

= 1− rnrm

Now, to show that∑ anrn

diverges, let sn = a1r1

+a2r2

+· · ·+anrn

. Since rn ↘ 0 (if we let tn =∑n

m=1 amand∑ an = t, then tn ↗ t and tn−1 + rn = t, so rn = t− tn−1 ↘ 0), given a positive integer M ,there exists an another positive integerN withN > M+1, such that n ≥ N implies rn < rM+1

2 ,or rnrM+1

< 12 . It follows that

sN − sM =aM+1

rM+1+aM+2

rM+2+ · · ·+ aN

rN> 1− rN

rM+1>

1

2

and then the sequence {sn} is not Cauchy. So {sn} diverges, i.e.,∑ anrn

diverges.—–(b) First, the inequality follows by writingan√rn

=rn − rn+1√

rn

=

√rn +

√rn+1√

rn(√rn −

√rn+1)

<

√rn +

√rn√

rn(√rn −

√rn+1)

= 2 (√rn −

√rn+1)


AoPS Community Chapter 3 Selected Exercises (Rudin)Now, to show that∑ an√

rnconverges, observe that

n∑k=1

ak√rk

< 2n∑k=1

(√rk −

√rk+1

)= 2 (

√r1 −

√rn+1)

Since rn ↘ 0, the r.h.s. of the above inequality converges to 2√r1 as n→∞. So we concludethat∑ an√

rnconverges.

13. Prove that the Cauchy product of two absolutely convergent series converges absolutely.—–Let ∑ an and ∑ bn be two absolutely convergent series, and ∑ cn be the Cauchy product ofthe two given series. Note that

|cn| =

∣∣∣∣∣n∑k=0

akbn−k

∣∣∣∣∣ ≤n∑k=0

|ak| |bn−k|

Since both∑ an and∑ bn converge absolutely, if we let dn =∑n

k=0 |ak| |bn−k|, then by Theorem3.50,∑ dn converges. Since |cn| ≤ dn, by the comparison test,∑ |cn| converges.14. If {sn} is a complex sequence, define its arithmetic means σn by

σn =s0 + s1 + · · ·+ sn

n+ 1(n = 0, 1, 2, . . .)

(a) If lim sn = s, prove that limσn = s.(b) Construct a sequence {sn} which does not converges, although limσn = 0.(c) Can it happen that sn > 0 for all n and that lim sup sn =∞, although limσn = 0?(d) Put an = sn − sn−1, for n ≥ 1. Show thatsn − σn =

1

n+ 1

n∑k=1

kak

Assume that lim(nan) = 0 and that {σn} converges. Prove that {sn} converges. [This gives aconverse of (a), but under the additional assumption that nan → 0.](e) Derive the last conclusion from a weaker hypothesis: Assume M <∞, |nan| ≤M for all n,and limσn = σ. Prove that lim sn = σ, by completing the following outline:If m < n, thensn − σn =

m+ 1

n−m(σn − σm) +

1

n−m

n∑i=m+1

(sn − si)

For these i,|sn − si| ≤

(n− i)Mi+ 1

≤ (n−m− 1)M

m+ 2


AoPS Community Chapter 3 Selected Exercises (Rudin)Fix ε > 0 and associate with each n the integer m that satisfies

m ≤ n− ε1 + ε

< m+ 1

Then (m+ 1)/(n−m) ≤ 1/ε and |sn − si| < Mε. Hencelim supn→∞

|sn − σ| ≤Mε

Since ε was arbitrary, lim sn = σ.—–(a) Given ε > 0, there exists a positive integer N such that n ≥ N implies |sn − s| < ε. Also,there exists a positive integer N ′ such that n ≥ N ′ implies∣∣∣∣s0 + s1 + · · ·+ sN−1 −Ns

n+ 1

∣∣∣∣ < ε

So n ≥ max{N,N ′} implies|σn − s| =

∣∣∣∣s0 + s1 + · · ·+ snn+ 1

− s∣∣∣∣

≤∣∣∣∣s0 + s1 + · · ·+ sN−1 −Ns

n+ 1

∣∣∣∣+

∣∣∣∣(sN − s) + (sN+1 − s) + · · ·+ (sn − s)n+ 1

∣∣∣∣< ε+

1

n+ 1(|sN+1 − s|+ |sN+2 − s|+ · · ·+ |sn − s|)

< ε+n−N + 1

n+ 1· ε

< 2ε

Hence limn→∞ σn = s.—–(b) For example, let sn = (−1)n for n = 0, 1, 2, . . ..—–(c) For example, let

sn =

3√n if n is a perfect cube;

1

2nif otherwise

Then clearly lim supn→∞ sn = ∞. However, given n, if we let k3 be the largest perfect cubesatisfying k3 ≤ n, thenσn ≤

2 + (1 + 2 + · · ·+ k)

k3 + 1=

2 + k2+k2

k3 + 1<k2 + k + 4

2k3→ 0


AoPS Community Chapter 3 Selected Exercises (Rudin)as k →∞. Hence limn→∞ σn = 0.—–(d) Observe that for m < n,

sn − sm = (sn − sn−1) + (sn−1 − sn−2) + · · ·+ (sm+1 − sm)

= an + an−1 + · · ·+ am+1

=n∑

k=m+1

ak

Thussn − σn = sn −

s0 + s1 + · · ·+ snn+ 1

=1

n+ 1[(sn − s0) + (sn − s1) + · · ·+ (sn − sn−1)]

=1

n+ 1

(n∑k=1

ak +n∑k=2

ak + · · ·+n∑

k=n

ak

)

=1

n+ 1

n∑k=1

kak

Now, assume that limn→∞(nan) = 0 and that {σn} converges, we want to show {sn} alsoconverges. By the previous result,sn = σn +

1

n+ 1

n∑k=1

kak

Also, if we let s′0 = 0 and s′n = nan for n = 1, 2, . . ., and let σ′n =s′0+s

′1+···+s′nn+1 . Then limn→∞ s

′n =

0, and then apply part (a), limn→∞ σ′n = 0. So

limn→∞

sn = limn→∞

(σn + σ′n

)= lim

n→∞σn + lim

n→∞σ′n = lim

n→∞σn

and we conclude that {sn} converges.—–(e) To complete the details of the outline, we prove the following three facts.(i) We show thatsn − σn =

m+ 1

n−m(σn − σm) +

1

n−m

n∑i=m+1

(sn − si)


AoPS Community Chapter 3 Selected Exercises (Rudin)Since

1

n−m

n∑i=m+1

(sn − si) = sn −1

n−m

n∑i=m+1

si

= sn −(n+ 1)σn − (m+ 1)σm

n−m

= sn −(n−m)σn − (m+ 1)σm + (m+ 1)σn

n−m

= sn −m+ 1

n−m(σn − σm)− σn

the result follows.(ii) We show that, for these i,|sn − si| ≤

(n− i)Mi+ 1

≤ (n−m− 1)M

m+ 2

We can directly write|sn − si| = |(sn − sn−1) + (sn−1 − sn−2) + · · ·+ (si+1 − si)|

= |an + an−1 + · · ·+ ai+1|≤ |an|+ |an−1|+ · · ·+ |ai+1|

≤ M

n+

M

n− 1+ · · ·+ M

i+ 1

≤ (n− i)Mi+ 1

and the result follows by taking i = m+ 1.(iii) To show that m+ 1

n−m≤ 1

ε, observe that m ≤ n− ε

1 + εimplies

m+ 1 ≤ (n− ε) + (1 + ε)

1 + ε=n+ 1

1 + ε

andn−m ≥ n− n− ε

1 + ε=n+ εn− n+ ε

1 + ε=ε(n+ 1)

1 + ε

Thus the result follows. Next, to show that |sn − si| < Mε, observe that n− ε1 + ε

< m+ 1 impliesn−m− 1 < n− n− ε

1 + ε=n+ εn− n+ ε

1 + ε=ε(n+ 1)

1 + ε

andm+ 2 = (m+ 1) + 1 >

n− ε1 + ε

+ 1 =n+ 1

1 + ε

Thus by (ii), the result follows.


AoPS Community Chapter 3 Selected Exercises (Rudin)16. Fix a positive number α. Choose x1 > √α, and define x2, x3, x4, . . ., by the recursion formula

xn+1 =1

2

(xn +

α

xn

)(a) Prove that {xn} decreases monotonically and that limxn =

√α.(b) Put εn = xn −

√α, and show that

εn+1 =ε2n

2xn<

ε2n2√α

so that, setting β = 2√α,

εn+1 < β

(ε1β

)2n

(n = 1, 2, 3, . . .)

(c) This is a good algorithm for computing square roots, since the recursion formula is simpleand the convergence is extremely rapid. For example, if α = 3 and x1 = 2, show that ε1/β < 110and that therefore

ε5 < 4 · 10−16, ε6 < 4 · 10−32

—–(a) First, we show that {xn} is decreasing, and that xn > √α for n = 1, 2, . . .. Starting x1 > √α,and suppose xn > √α holds for some n ≥ 1. Then since 1

xn< 1√

α, we have α

xn<√α < xn, and

xn+1 =1

2

(xn +

α

xn

)<

1

2(xn + xn) = xn

On the other hand, by the arithmetic-geometric inequality, xn+1 >

√xn

(αxn

)=√α, where the

strict inequality symbol ”>” follows by the fact αxn6= xn (since α

xn< xn). Next, we show that

limn→∞ xn =√α. By the above result, {xn} is monotonically decreasing with lower bound√α.This means that limn→∞ xn = L for some real L, and then L = 1

2

(L+ α

L

) or L =√α.—–(b) Expand the following two values

εn+1 = xn+1 −√α =

1

2

(xn +

α

xn

)−√α =

xn2

+α

2xn−√α

andε2n

2xn=

(xn −√α)

2

2xn=x2n + α− 2xn

√α

2xn=xn2

+α

2xn−√α


AoPS Community Chapter 3 Selected Exercises (Rudin)we obtain εn+1 = ε2n

2xn< ε2n

2√α

. Setting β = 2√α, then recursively

εn+1 <1

β· ε2n

<1

β · β2· ε4n−1

<1

β · β2 · β4· ε8n−2

...<

1

β · β2 · β4 · · ·β2n−1 · ε2n

1

= β

(ε1β

)2n

—–(c) Note that β = 2√

3 and ε1 = 2−√

3, computeε1β

=2−√

3

2√

3=

1√3− 1

2

Since 1√3

+ 12 >

12 + 1

2 = 1, andε1β

(1√3

+1

2

)=

(1√3− 1

2

)(1√3

+1

2

)=

1

12<

1

10

we conclude that ε1β < 1

10 . Therefore by part (b),ε5 < β

(ε1β

)24

< 2√

3 · 10−16 < 4 · 10−16

and similarly ε6 < 4 · 10−32.17. Fix α > 1. Take x1 > √α, and define

xn+1 =α+ xn1 + xn

= xn +α− x2n1 + xn

(a) Prove that x1 > x3 > x5 > · · · .(b) Prove that x2 < x4 < x6 < · · · .(c) Prove that limxn =√α.(d) Compare the rapidity of convergence of this process with the one described in Exercise 16.—–


AoPS Community Chapter 3 Selected Exercises (Rudin)(a) Observe that

xn+2 =α+ xn+1

1 + xn+1

=α+

α+ xn1 + xn

1 +α+ xn1 + xn

=2α+ (α+ 1)xnα+ 1 + 2xn

= xn + 2

(α− x2n

α+ 1 + 2xn

)Note that x1 > √α. Now suppose xn > √α for some odd number n, then clearly xn+2 < xn.On the other hand, since√α > 1, we have α+ 1 > 2

√α. It follows that

xn+2 = xn + 2

(α− x2n

α+ 1 + 2xn

)> xn + 2

(α− x2n

2√α+ 2xn

)= xn +

(√α+ xn) (

√α− xn)√

α+ xn

=√α

So we conclude that x1 > x3 > x5 > · · · .—–(b) First, since α− x21 < 0 and√α > 1, we havex2 = x1 +

α− x211 + x1

< x1 +α− x21√α+ x1

=√α

Now suppose xn < √α for some even number n ≥ 2, then clearly xn+2 > xn [by the first equal-ity in part (a)]. On the other hand, use a similar argument to that in part (a), xn+2 <√α. So weconclude that x2 < x4 < x6 < · · · .—–(c) By part (a) and (b), the sequences {x2k−1} and {x2k} (k = 1, 2, . . .) are monotonically de-creasing and increasing, respectively. Also,√α serves as a lower and upper bound of {x2k−1}and {x2k}, respectively. So we let limk→∞ x2k−1 = L1 and limk→∞ x2k = L2. Note that we mayassume

L1 = L1 + 2

(α− L2

1

α+ 1 + 2L1

)or L1 =

√α

and similarly L2 =√α. It is easy to conclude that limn→∞ xn =

√α.—–


AoPS Community Chapter 3 Selected Exercises (Rudin)(d) We imitate an analysis in Exercise 16(b). Put εn = |xn −

√α| and β =

√α−1

1+x2, then observe

that 1 < 1+x2 (since x2 = α+x11+x1

> 0) and 1+x2 ≤ 1+xn for all n [by part (a) and (b)]. It followsthatεn+1 =

∣∣xn+1 −√α∣∣

=

∣∣∣∣α+ xn1 + xn

−√α

∣∣∣∣=

∣∣∣∣xn −√α−√α (xn −√α)

1 + xn

∣∣∣∣=

∣∣∣∣1−√α1 + xn

∣∣∣∣ εn≤ βεn

and thereforeεn ≤ βεn−1 ≤ β2εn−2 ≤ · · · ≤ βn−1ε1

So roughly the speed of convergence of the process is not more rapid than the one in Exercise16.18. Replace the recursion formula of Exercise 16 by

xn+1 =p− 1

pxn +

α

px−p+1n

where p is a fixed positive integer, and describe the behavior of the resulting sequence {xn}.—–Observe that if the limit of {xn} exists, saying L, then

L =p− 1

pL+

α

pL−p+1 or L = α

1p

Now, we take x1 > α1p , and suppose xn > α

1p for some positive integer n, then x−pn < α−1 and

xn+1 <p− 1

pxn +

α

pα−1xn = xn

On the other hand, by the arithmetic-geometric inequality,xn+1 >

p

√xp−1n αx−p+1

n = α1p

where the strict inequality symbol ”>” follows by the fact xn 6= αx−p+1n (since x−pn < α−1

or αx−pn < 1). So we conclude that {xn} decreases monotonically with lower bound α 1p , and

hence {xn} converges to L = α1p , i.e., limn→∞ xn = α

1p .

Remark. Exercise 16 is a special case that p = 2.


AoPS Community Chapter 3 Selected Exercises (Rudin)19. Associate to each sequence a = {an}, in which an is 0 or 2, the real number

x(a) =

∞∑n=1

αn3n

Prove that the set of all x(a) is precisely the Cantor set described in Sec. 2.44.—–We will use the notationsE1, E2, E3, . . . and P that can be found in Sec. 2.44 (p.41). Let x(a) =∑∞

n=1an3n , where an is 0 or 2. Consider the first term of x(a), i.e., a13 , then either a13 = 0 or a13 = 2

3 .It is clear that the first term of x(a) is located at the lower endpoint of one of the two intervalsin E1, i.e., a13 ∈ P . Now, suppose that the partial sum ∑mn=1

an3n of x(a) is located at the lowerendpoint of one of the 2m intervals in Em, saying Im. Then ∑mn=1

an3n ∈ P , and the partial sum∑m+1

n=1an3n must be located at the lower endpoint of one of the two intervals in Im ∩ Em+1, i.e.,∑m+1

n=1an3n ∈ P . Continue this process, we see that either x(a) ∈ P or x(a) is a limit point of

P , and hence x(a) ∈ P (since P is closed). To show the converse, use the fact that (as alsoshown in the text) every point of P can be written as a limit of a sequence {xm} of lowerendpoints of some interval Im ⊂ Em (actually we can write xm =∑m

n=1an3n , where an is 0 or 2),for m = 1, 2, . . ..

20. Suppose {pn} is a Cauchy sequence in a metric space X , and some subsequence {pni} con-verges to a point p ∈ X. Prove that the full sequence {pn} converges to p.—–Given ε > 0. Since {pni} converges to p, there is a positive integer N1 such that i ≥ N1 impliesd(pni , p) < ε. Since {pn} is Cauchy, there is a positive integer N2 such that d(pm, pn) < ε for allm ≥ N2 and n ≥ N2. Choose N ≥ N2 such that N = ni for some i ≥ N1, then for all n ≥ N2,

d(pn, p) ≤ d(pn, pN ) + d(pN , p) < ε+ ε = 2ε

Hence {pn} converges to p.21. Prove the following analogue of Theorem 3.10(b): If {En} is a sequence of closed nonemptyand bounded sets in a complete metric space X , if En ⊃ En+1, and if

limn→∞

diamEn = 0

then⋂∞1 En consists of exactly one point.—–Denote E =

⋂∞n=1En. Suppose E consists of at least two points, then diamE > 0. But since

E ⊂ En for all n, we have diamE ≤ diamEn for all n, which contradicts the assumption thatlimn→∞ diamEn = 0. Next, we show thatE 6= ∅. SinceEn 6= ∅ for all n, we can choose a pointxn ∈ En for each n, and form the sequence {xn}. We claim that {xn} is Cauchy.Proof of the claim. Given ε > 0, then since limn→∞ diamEn = 0, there exists a positive integer


AoPS Community Chapter 3 Selected Exercises (Rudin)N such that n ≥ N implies diamEn < ε. For any m ≥ N and n ≥ N , by assumption, xm ∈Em ⊂ EN and xn ∈ En ⊂ EN ,. So d(xm, xn) ≤ diamEN < ε. �Now, since X is complete and {xn} is Cauchy, we have limn→∞ xn = x for some x ∈ X. Toshow that such x ∈ E, suppose not, then x /∈ EM or x ∈ EcM for some positive integerM . SinceEM is closed, EcM is open, and there exists a δ > 0 such that Nδ(x) ⊂ EcM , or Nδ(x)∩EM = ∅.Also, note that xn ∈ En ⊂ EM for all n ≥ M . It follows that d(x, xn) ≥ δ for all n ≥ M ,contradicting the assumption limn→∞ xn = x.

22. Suppose X is a nonempty complete metric space, and {Gn} is a sequence of dense opensubsets of X. Prove Baire’s theorem, namely, that ⋂∞1 Gn is not empty. (In fact, it is densein X.) Hint: Find a shrinking sequence of neighborhoods En such that En ⊂ Gn, and applyExercise 21.—–Since X 6= ∅, we begin with some point x0 ∈ X and given r > 0. Since G1 is dense andopen, the set Nr(x0) ∩G1 is nonempty and open. Choose x1 ∈ N1(x0) ∩G1 and there exists aneighborhood E1 of x1 such that

E1 ⊂ Nr(x0) ∩G1 and diamE1 ≤ r

Having chosen nonempty neighborhoods E1, E2, . . . , En−1, such that Ei ⊂ Gi, Ei+1 ⊂ Ei, anddiamEi ≤ 21−ir. SinceGn is dense and open, the setEn−1∩Gn is nonempty and open. Choosexn ∈ En−1 ∩Gn and there exists a neighborhood En of xn such that

En ⊂ En−1 ∩Gn and diamEn ≤ 21−nr

Continue this process, we may construct the desired shrinking sequence {En} such that En ⊂Gn for all n, and limn→∞ diamEn = 0. Then by Exercise 21, ⋂∞n=1 En consists of one point,saying x. Observe that⋂∞n=1 En ⊂

⋂∞n=1Gn, we have x ∈ ⋂∞n=1Gn and hence⋂∞n=1Gn 6= ∅. Infact, ⋂∞n=1Gn is dense in X , because x ∈ E1 ⊂ Nr(x0), where x0 ∈ X and r > 0 are arbitrarilygiven.

23. Suppose {pn} and {qn} are Cauchy sequences in a metric space X. Show that the sequence{d(pn, qn)} converges. Hint: For any m, n,

d(pn, qn) ≤ d(pn, qm) + d(pm, qm) + d(pm, qn)

it follows that|d(pn, qn)− d(pm, qm)|

is small if m and n are large.—–Given ε > 0, then since {pn} and {qn} are Cauchy, there exist positive integers N1 and N2such that m,n ≥ N1 implies d(pm, pn) < ε, and m,n ≥ N2 implies d(qm, qn) < ε. Put N =


AoPS Community Chapter 3 Selected Exercises (Rudin)max{N1, N2}, then by the hint, for m,n ≥ N ,

d(pn, qn) ≤ d(pn, qm) + d(pm, qm) + d(pm, qn) < d(pm, qm) + 2ε

which follows that |d(pn, qn)− d(pm, qm)| < 2ε. Thus {d(pn, qn)} is Cauchy in R1, and hence{d(pn, qn)} converges by Theorem 3.11(c).

24. Let X be a metric space.(a) Call two Cauchy sequences {pn}, {qn} in X equivalent iflimn→∞

d(pn, qn) = 0

Prove that this is a equivalence relation.(b) Let X∗ be the set of all equivalence classes so obtained. If P ∈ X∗, Q ∈ X∗, {pn} ∈ P ,{qn} ∈ Q, define

∆(P,Q) = limn→∞

d(pn, qn)

by Exercise 23, this limit exists. Show that the number ∆(P,Q) is unchanged if {pn} and {qn}are replaced by equivalent sequences, and hence that ∆ is a distance function in X∗.(c) Prove that the resulting metric space X∗ is complete.(d) For each p ∈ X , there is a Cauchy sequence all of whose terms are p; let Pp be the elementof X∗ which contains this sequence. Prove that∆(Pp, Pq) = d(p, q)

for all p, q ∈ X. In other words, the mapping ϕ defined by ϕ(p) = Pp is an isometry (i.e., adistance-preserving mapping) of X into X∗.(e) Prove that ϕ(X) is dense in X∗, and that ϕ(X) = X∗ if X is complete. By (d), we mayidentify X and ϕ(X) and thus regard X as embedded in the complete metric space X∗. Wecall X∗ the completion of X.—–(a) Let {pn}, {qn}, and {rn} be Cauchy sequences in X.(Reflexivity) Since d(pn, pn) = 0 for all n, we have limn→∞ d(pn, pn) = 0. So {pn} is equivalentto itself.(Symmetry) If {pn} is equivalent to {qn}, then since d(pn, qn) = d(qn, pn) for all n, we have

limn→∞

d(qn, pn) = limn→∞

d(pn, qn) = 0

i.e., {qn} is equivalent to {pn}.(Transitivity) If {pn} is equivalent to {qn}, and if {qn} is equivalent to {rn}. Then we have limn→∞ d(pn, qn) =0 and limn→∞ d(qn, rn) = 0. Also, by the triangular inequality, d(pn, rn) ≤ d(pn, qn) + d(qn, rn)


AoPS Community Chapter 3 Selected Exercises (Rudin)for all n. It follows by the comparison test that

limn→∞

d(pn, rn) ≤ limn→∞

[d(pn, qn) + d(qn, rn)]

= limn→∞

d(pn, qn) + limn→∞

d(qn, rn)

= 0 + 0

= 0

or limn→∞ d(pn, rn) = 0. So {pn} is equivalent to {rn}.—–(b) Let {pn}, {p′n} ∈ P and {qn}, {q′n} ∈ Q, where P,Q ∈ X∗. Then by definition,limn→∞

d(pn, p′n) = lim

n→∞d(qn, q

′n) = 0

Also, by the triangular inwquality,d(p′n, q

′n) ≤ d(p′n, pn) + d(pn, qn) + d(qn, q

′n)

which follows thatlimn→∞

d(p′n, q′n) ≤ lim

n→∞

[d(p′n, pn) + d(pn, qn) + d(qn, q

′n)]

= limn→∞

d(p′n, pn) + limn→∞

d(pn, qn) + limn→∞

d(qn, q′n)

= limn→∞

d(pn, qn)

and by symmetry limn→∞ d(pn, qn) ≤ limn→∞ d(p′n, q′n). So limn→∞ d(pn, qn) = limn→∞ d(p′n, q

′n).Now, we show that ∆ is a distance function in X∗.(1) By definition, it is clear that ∆(P,Q) ≥ 0 for all P,Q ∈ X∗. Also, ∆(P,Q) = 0 if and only if

limn→∞ d(pn, qn) = 0 for all {pn} ∈ P and {qn} ∈ Q, if and only if {pn} is equivalent to {qn} forall {pn} ∈ P and {qn} ∈ Q, if and only if P = Q.(2) The equality ∆(P,Q) = ∆(Q,P ) holds simply by definition.(3) Given P,Q,R ∈ X∗, choose {pn} ∈ P , {qn} ∈ Q, and {rn} ∈ R. Then by definition and bythe triangular inequality (w.r.t. d),∆(P,Q) = lim

n→∞d(pn, qn)

≤ limn→∞

[d(pn, rn) + d(rn, qn)]

= limn→∞

d(pn, rn) + limn→∞

d(rn, qn)

= ∆(P,R) + ∆(R,Q)

i.e., the triangular inequality (w.r.t. ∆) holds.By checking (1) (3), we conclude that ∆ is a distance function in X∗.—–


AoPS Community Chapter 3 Selected Exercises (Rudin)(c) Let {Pn} be a Cauchy sequence in X∗, and choose {pnk} ∈ Pn for n = 1, 2, . . .. For eachn, there exists a positive integer Kn such that k, l ≥ Kn implies d(pnk , p

nl ) < 2−n. Define

pn = pnKnfor n = 1, 2, . . ., we show that {pn} is Cauchy in X. Given ε > 0, then since {Pn} isCauchy in X∗, there exists a positive integer N such that m,n ≥ N implies ∆(Pm, Pn) < ε, or

limk→∞ d(pmk , pnk) < ε. LetN ′ be a positive integer such that 2−N

′< ε, thenm,n ≥ max{N,N ′}implies

d(pm, pn) = d(pmKm, pnKn

)

≤ d(pmKm, pmk ) + d(pmk , p

nk) + d(pnk , p

nKn

)

< 2−m + ε+ 2−n

< 3ε

where the integer k can be chosen sufficiently large such that k ≥ max{Km,Kn} and d(pmk , pnk) <

ε. Hence {pn} is Cauchy in X. Let P ∈ X∗ be such that {pn} ∈ P , it suffices to show thatPn → P , which completes the proof. Given ε > 0, put n ≥ max{N,N ′}, k ≥ max{N,N ′,Kn},and m ≥ max{Kn,Kk} sufficiently large such that d(pnm, p

km) < ε. We have

d(pnk , pk) = d(pnk , pkKk

)

≤ d(pnk , pnm) + d(pnm, p

km) + d(pkm, p

kKk

)

< 2−n + ε+ 2−k

< 3ε

which follows that limk→∞ d(pnk , pk) ≤ 3ε, or ∆(Pn, P ) ≤ 3ε. Hence Pn → P .—–(d) The result is trivial [by part (b)].—–(e) (i) Let P ∈ X∗ and {pn} ∈ P , and given ε > 0. Since {pn} is Cauchy, there exists a positiveinteger N such that m,n ≥ N implies d(pm, pn) < ε. It follows that∆(P,ϕ(pN )) = ∆(P, PpN ) = lim

n→∞d(pn, pN ) ≤ ε < 2ε

i.e., the neighborhood of P with radius 2ε contains an element ϕ(pN ) ∈ ϕ(X). So we concludethat ϕ(X) is dense in X∗.(ii) If X is complete, let P ∈ X∗ and {pn} ∈ P . Since {pn} is Cauchy, pn → p for some p ∈ X ,i.e., we have∆(P, Pp) = lim

n→∞d(pn, p) = 0

So P = Pp, and hence X∗ = ϕ(X).25. Let X be the metric space whose points are the rational numbers, with the metric d(x, y) =

|x− y|. What is the completion of this space? (Compare Exercise 24.)—–


AoPS Community Chapter 3 Selected Exercises (Rudin)By Theorem 3.11(c), every rational Cauchy sequence converges to a real number. On the otherhand, for every x ∈ R1, there exists a rational Cauchy sequence {pn} such that limn→∞ pn = x.So the completion of X should be ϕ(R1).

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AoPS Community Chapter 3 Selected Exercises (Rudin)

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