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Considertheseriesk=1(-1)k+1akandk=1(-1)kakwhere(i){ak}isasequenceofpositivenumber(ii)thesequence{ak}isstrictlydecreasing(iii)limkak=0

Find an example of a divergent series of the form

(a) that satisfies conditions (i) and (iii), but not condition (ii);

(b) that satisfies conditions (i) and (ii), but not condition (iii).

Short Answer

Expert verified

(i)(i)Theseriesk=1(-1)k+1aksatisfyingthegivenconditionisk=1(-1)k+12k.(ii),Theseriesk=1(-1)k+1aksatisfyingthegivenconditionisk=1(-1)k+1k2k-1.

Step by step solution

01

Step 1. Given

Considertheseriesk=1(-1)k+1akandk=1(-1)kakwhere(i){ak}isasequenceofpositivenumber(ii)thesequence{ak}isstrictlydecreasing(iii)limkak=0

02

Step 2. Hypothesis of alternating series

Itstatesthatif{ak}isasequenceofpositivenumberwithak+!<akforeveryk1andlimkak=0thenalternatingseriesk=1(-1)k+1akandk=1(-1)kakbothconvergeisak+1<ak

03

Part(a) Step 3. Explanation

Considertheseriesk=1(-1)k+1ak=k=1(-1)k+12kThesequenceak=12kisapositivetermsThevalueoflimkak=limk1k=0Buttheseriesk=1(-1)k+1ak=k=1(-1)k+12kisanincreasingseries.Thus,theseriesk=1(-1)k+1aksatisfyingthegivenconditionisk=1(-1)k+12k.

04

Part(b) Step 4. Explanation

Considertheseriesk=1(-1)k+1ak=k=1(-1)k+1k2k-1Thesequenceak=k2k-1isapositivetermsThevalueoflimkak=limkk2k-1=12Thus,theseriesk=1(-1)k+1aksatisfyingthegivenconditionisk=1(-1)k+1k2k-1.

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Most popular questions from this chapter

Prove Theorem 7.25. That is, show that the series k=1akandk=Makeither both converge or both diverge. In addition, show that if k=Makconverges to L, thenk=1akconverges tolocalid="1652718360109" a1+a2+a3+....+aM-1+L.

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True/False:

Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: If ak0, then k=1akconverges.

(b) True or False: If k=1akconverges, then ak0.

(c) True or False: The improper integral 1f(x)dxconverges if and only if the series k=1f(k)converges.

(d) True or False: The harmonic series converges.

(e) True or False: If p>1, the series k=1k-pconverges.

(f) True or False: If f(x)0as x, then k=1f(k) converges.

(g) True or False: If k=1f(k)converges, then f(x)0as x.

(h) True or False: If k=1ak=Land {Sn}is the sequence of partial sums for the series, then the sequence of remainders {L-Sn}converges to 0.

Let αbe any real number. Show that there is a rearrangement of the terms of the alternating harmonic series that converges to α. (Hint: Argue that if you add up some finite number of the terms of k=112k1, the sum will be greater than α. Then argue that, by adding in some other finite number of the terms of

k=112k , you can get the sum to be less than α. By alternately adding terms from these two divergent series as described in the preceding two steps, explain why the sequence of partial sums you are constructing will converge to α.)

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