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Math 431 - Real Analysis I

Solutions to Homework due December 5

Question 1.

(a) Letak;bk0 for allk. Show that if1X k=0a kconverges andbkis a bounded sequence, then1X k=0a kbk converges as well. (b) Find a counterexample to above statement if the hypothesis \ak;bk0" is removed.

Solution 1.

(a) Sincebkis a bounded sequence, we have thatjbkj Mfor someM. Sincebk0< we have thatbkM for allk. Sinceak0, we have that

0akbkMak:

Notice that sinceP1

k=0akconverges, thenP1 k=0Makconverges as well. Thus, by the Comparison

Test, we have that1X

k=0a kbk converges as well. (b) Consider the convergent sequence 1X k=1a k=1X k=1(1)k1k and the bounded sequencebk= (1)k. Notice that the sequenceakbk=1k , but 1 X k=11k diverges.Question 2.Show that1X k=21k(lnk)p converges if and only ifp >1.

Solution 2.We will show thatP1

k=21k(lnk)pconverges if and only ifp >1 by using the integral test.

Consider the function

f(x) =1x(lnx)p dened forx2. Notice thatf(x)0 for allx2 and that

01x(lnx)p1x

Thus, by the Squeeze Theorem, we have that lim

x!1f(x) = 0. Furthermore, notice that ifxy, then (ln(x))p(ln(y))p:Thus, we get thatx(lnx)py(lny)p. Thus, f(y) =1y(lny)p1x(lnx)p=f(x): 1

Thus,fis a decreasing function. So, we can apply the integral test. Notice that lettingu= lnxanddu=dxx

we have thatZ1

2dxx(lnx)p= limb!1Z

b

2dxx(lnx)p= limb!1Z

lnb ln2duu p:

Ifp= 1, then the above integral evaluates to

lim b!1lnjlnbj ln2 =1:

Ifp6= 1, then the above integral evaluates to

lim b!1b

1p1p+1p1:

This will converge if and only ifp <1.

Thus, by the integral test, our series converges if and only ifp >1.Question 3. (a) Give an example of a divergent series P1 k=1akfor whichP1 k=1a2kconverges. (b) Prove that ifak0 andP1 k=1akconverges, thenP1 k=1a2kconverges as well. (c) Find a counterexample to the above statement if the hypothesis \ak0" is removed.

Solution 3.

(a) Consider the divergent series 1X k=11k

Notice that

1X k=11k 2 converges. (b) Since P1 k=1akconverges, thenak!0 by the Divergence Test. Thus, there exists someNsuch that for alln > N,jan0j<1, and thusan<1. So, for alln > N,an<1 and thusa2n< an. Thus, by the

Generalized Comparison Test, sinceP1

k=1akconverges, thenP1 k=1a2kconverges as well. (c) Consider the alternating series 1X k=1(1)kpk which converges by the Alternating Series Test. Squaring the terms, we get that X k=11k diverges. 2 Question 4.In this question, we will show that if both 1 X k=0a

2kand1X

k=0b 2k converges, then 1X k=0a kbk converges absolutely. (a) Show that 2jakbkj a2k+b2kfor allk. (b) Use (a) to show that if P1 k=0a2kandP1 k=0b2kconverges, thenP1 k=0akbkconverges absolutely.

Solution 4.

(a) Notice that (akbk)20. Thus, we have thata2k2akbk+b2k0. Rewriting, we get that

2akbka2k+b2k:

Similarly, consider (ak+bk)20. Thus, we have thata2k+ 2akbk+b2k0. Re-writing, we have that (a2k+bk)22akbk: Putting these two inequalities together, we have that (a2k+bk)22akbka2k+b2k; which is equivalent to our desired result. (b) Since P1 k=0a2kandP1 k=0b2kconverges, their sum also converges and is, in fact, equal toP1 k=0a2k+b2k. Since 0<2jakbkj a2k+b2k, by the Comparison Theorem, we have that 1 X k=02jakbkj converges, and thus 1X k=0jakbkj converges. Thus, 1X k=0a kbk

converges absolutely.Question 5.Letanbe a sequence of non-zero real numbers such that the sequencean+1a

nof ratios is a constant sequence. Show thatP1 k=1akis a geometric series.

Solution 5.Let's assume thatan+1a

nis equal to the constant sequencer. Then, we have that for alln, a n+1a n=r 3 and thusan+1=ran. This gives a recursive denition ofan. So, assume thata0=c. Then,a1=ra0=cr. Similarlya2=ra1=cr2. Continuing in this way (using induction), one can see thatan=crn. Thus,P1 k=0ak=P1

k=0crk, a geometric series.In class, we learned of the following denition for thelimit superiorandlimit inferiorof a sequencexn.

Suppose a numberUhas the following two properties: For all" >0, there exists anNsuch that for alln > N,xn< U+"; and For all" >0 andm >0, there exists ann > msuch thatxn> U".

Then, limsupxn=U.

Similarly, suppose that a numberLhas the following two properties: For all" >0, there exists anNsuch that for alln > N,xn> L"; and For all" >0 andm >0, there exists ann > msuch thatxn< L+":

Then liminfxn=L.

Question 6.Letxnbe a sequence. If liminfxn=A= limsupxn, show thatxnconverges and that, in fact,xn!A. Solution 6.Let" >0. Since liminfxn=A, then there exists anNsuch that for alln > N,xn> A". Since limsupxn=A, then there exists anN+such that for alln > N+,xn< A+". Choose N= maxfN;N+g. Then, for alln > N, we have thatA" < xn< A+";which is equivalent to jxnAj< ". So,xn!A.Question 7.Consider the series1X k=1pk+ 1pk: (a) Show that pk+ 1pk=1pk+ 1 +pk (b) Use (a) to decide if the series converges or diverges.

Solution 7.

(a) If we take our term and multiply by a special form of 1, we get pk+ 1pkpk+ 1 +pkpk+ 1 +pk =k+ 1kpk+ 1 +pk =1pk+ 1 +pk 4 (b) We can now limit compare our terms to 1pk to get

1pk+1+pk

1pk =pkpk+ 1 +pk =1q k+1k + 1=12

Thus, since

1X k=11pk

diverges by thep-series test, we have that our original series diverges as well.Question 8.Consider the series1X

k=1a kwhere a k=2(1)k3 k (a) Compute ak+1a k for generalk. Use this to compute limsup a k+1a k and liminfa k+1a k Can you use the Ratio Test to reach a conclusion about the convergence or divergence of the series? (b) Computejakj1=kfor generalk. Use this to computequotesdbs_dbs7.pdfusesText_13

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