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Solution to Principles of Mathematical Analysis Chapter 3 Part A


Chapter 3 Numerical Sequences and Series

Exercise 1

(By ghostofgarborg) Assume sns. Choose ϵ>0, and let N be such that |sns|<ϵ when n>N. Then ||sn||s|||ssn|<ϵwhenever n>N. This implies convergence of |sn|.

The converse is not true. Let sn=(1)n. This sequence does not converge, even though |sn| does.


Exercise 2

(By ghostofgarborg) Note that (n2+nn)=n/(n2+n+n), and therefore
12+1n=n(n+n)+n<n2+nn<nn+n=12where (2+1/n)112. By theorem 3.19, limnn2+nn=12.


(By Dan kyp44 Whitman) We can calculate this limit directly:
limn(n2+nn)=limn(n2+nn)(n2+n+nn2+n+n)=limnn2+nn2n2+n+n=limnnn2+n+n=limn11+1/n+1=11+1=12,noting that clearly n2+n+n>0.


Exercise 3

(By ghostofgarborg) Note that if sn<2, sn+1=2+sn<4=2, so the sequence is bounded. Also note that if sn1>sn2, then sn=2+sn1>2+sn2=sn1, so the sequence is monotonically increasing. It therefore converges by thm 3.14.


Exercise 4

(By ghostofgarborg) We observe that if an=(2n1)/2n and bn=(2n11)/2n, then s2n+1=an and s2n=bn fulfills both the initial condition and the recursion, so that this is a closed form of the sequence. Note that an1, bn12, and that any subsequence of sn contains either a subsequence of an or a subsequence of bn. Consequently, any convergent subsequence converges to either 1 or 12. We get lim supnsn=1lim infnsn=12.


Exercise 5

(By ghostofgarborg) If lim supan= and lim supbn>, the result is trivial. If lim supan= and lim supbn<, then bn is bounded above by B and given a number N #{n:an+bn>N}#{n:an>NB}the latter of which is finite. Therefore lim sup(an+bn)==lim supan+lim supbn.

We can therefore assume that both lim supan and lim supbn are finite. There is a subsequence such that ank+bnklim sup(an+bn). By choosing this subsequence right, i.e. if necessary passing to another subsequence, we can make sure that ank is convergent. Then bnk=(ank+bnk)ank is convergent, so
lim supn(an+bn)=limk(ank+bnk)=limkank+limkbnklim supnan+lim supnbn.


Exercise 6

(By ghostofgarborg)

(a) Since the series telescopes, Sn=i=0n(i+1i)=n+1. This shows that it diverges to .

(b) We havean=1n(n+1+n)<1nn.Converges by the comparison test (thm. 3.25).

(c) We have lim supn|an|n=lim supn(nn1)=11=0Converges by the root test.

(d) When |z|1, |11+zn|11+|z|n12violating the necessary condition that an0. In this case, an
is divergent. When |z|>1, |11+zn|1|z|n11|z|n|z|n1=(|z||z|1)1|z|n and an converges by comparison with the geometric series |z||z|11|z|n.


Exercise 7

(By ghostofgarborg) Assume an=C. We know that 1n2 converges. Let D be the number it converges to. Note that Sn=i=0nai/i is monotonic. By the Cauchy-Schwarz inequality i=1naii(i=1nai)12(i=0n1i2)12<CDThis implies that ann is bounded, hence converges.


Exercise 8

(By ghostofgarborg) We first note that thm 3.42 holds for bn a monotonously increasing sequence whose limit is 0 as well, since (bn) then fulfills the criteria of the theorem, and anbn=an(bn).

If an converges, the partial sums form a bounded sequence. If bn is monotonic and bounded it converges to a number B, and we get that anbn=an(bnB)+BanThe first sum on the right hand side converges by thm 3.42 and the observation above. The second sum converges because an does. Consequently the left hand side converges.


Exercise 9

(By ghostofgarborg)

(a) We have1R=limn(n+1)3n3=limn(1+3n+3n2+1n3)=1So R=1.

(b) We have 1R=limn2n+1(n+1)!2nn!=limn2n+1=0So R=.

(c) We have 1R=limn2nn2n=limn2(nn)2=2 So R=12.

(d) We have 1R=limnn33nn=limn(nn)33=13 So R=3.


Exercise 10

(By ghostofgarborg) If |z|1 and an0 for an infinite number of values of n, then |anzn||an|, and does not tend to 0. This makes the series divergent when |z|1.


Exercise 11

(By ghostofgarborg)

(a) If an1, then an1+anan2. If an>1, then an1+an>12. Therefore an1+an12min(an,1) If {n:an>1} is infinite, this clearly diverges. Otherwise, there is an N such that n>N implies an1, in which case the series diverges by comparison to 12an.

(b) Since Sn is monotonically increasing, whenever jk an+jSn+jan+jSn+k.Consequently, aN+1SN+1++aN+kSN+kaN+1SN+k++a+N+kSN+k=aN+1++aN+kSN+k=1SNSN+kSince Sn is increasing and unbounded, it is possible to choose k such that SN/SN+k is arbitrarily close to 0. This shows that anSn is not Cauchy, and therefore not convergent.

(c) Since Sn is monotonically increasing, we have that Sn2Sn1Sn, so we can deduce that anSn2SnSn1Sn1Sn=1Sn11Sn Since 1/Sn is bounded n=1NanSn2a1S12+n=2N(1Sn11Sn)a1S12+1S11SN so an/Sn2 is monotonic and bounded, hence convergent.

(d) Since an1+n2an1n2 an/(1+n2an) converges. The series an/(1+nan), however, might converge or diverge. Let an=1n, and it is clear that it diverges. Let an=1 whenever n is a square and an=2n otherwise. This series clearly diverges, since the terms do not tend to 0 as n. Then n=1Nan1+nann=1N1n2+n=1N12n and the series therefore converges.


Exercise 12

(By ghostofgarborg)

(a) Since rn is monotonically decreasing amrm++anrn>amrm++an1rm=am++an1rm=rmrnrm=1rnrm Since rn0, given any M we can find an N>M such that 1RNRM is arbitrarily close to 1. This implies that an/rn is not Cauchy, hence not convergent.

(b) We have anrn=an(1+rn+1rn)rn+rn+1<2anrn+rn+1=2(rnrn+1)rn+rn+1=2(rnrn+1)Since n=1N2(rnrn+1)=2(r1rN+1)<2r1the series converges by the comparison test.


Exercise 13

(By ghostofgarborg) Since |an| and |bn| converge, so does their Cauchy product Cn. Let cn be the Cauchy product of an and bn. Then |cn|Cn, and as it is majorized by a convergent series, |cn| converges . I.e. cn converges absolutely.


Exercise 14

(By ghostofgarborg)

(a) Choose ϵ>0 and N such that |snsm|<ϵ/2 for all n,m>N. Then choose M such that |(s0s)++(sN1s)M|<ϵ2 Then whenever n>max(M,N) |(s0s)++(sns)n+1||(s0s)++(sN1s)n+1|+|(SNs)++(Sns)n+1|<ϵ2+ϵ2=ϵThis shows that (σns) converges to 0, i.e. that σns.

(b) Let sn=(1)n. Then sn does not converge, but σn=1+(1)n2(n+1) converges to 0.

(c) Yes. Let an=12n, b2n=n, and bn=0 when n is not a power of 2.
Let sn=an+bn. Then sn>0, and the series is unbounded, so lim supsn=. However0<σn=1n+1i=0n12i+1n+1i=0log2(n)i<2n+log2(n)2nwhere the right hand side tends to 0. This shows that σn0.

(d) Note thatk=1nk(sksk1) = k=1nkskk=0n1(k+1)sk = nsnk=0n1sk = (n+1)snk=0nskso that
snσn=1n+1k=0nkak Note that the right hand side is the nth arithmetic mean of the sequence nan, which by assumption and our result in (a) converges. Therefore snσn is convergent, and since σn is convergent by assumption, (snσn)+σn=sn is convergent.

(e) We have snσn=snk=0nskn+1=snk=0n(1nmm+1(n+1)(nm))sk=k=0m((m+1)sk(n+1)(nm)sknm)+k=m+1n((m+1)sk(n+1)(nm)+snsknm)=m+1nm(k=0nskn+1k=0mskm+1)+1nmk=m+1n(snsk) which is the desired equality. By assumption |snsn1|<M/n, so that |snst|=|(snsn1)+(sn1sn2)++(si+1si)|Mn++Mi+1(ni)Mi+1The rest of the argument is covered in sufficient detail in the text.

Baby Rudin 数学分析原理完整第三章习题解答


Linearity

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