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Chapter 7: Q. 30 (page 625)

Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges.

∑k=1∞ (3k−1)−2/3

Short Answer

Expert verified

Ans: The series∑k=1∞ (3k−1)−2/3is divergent.

Step by step solution

01

Step 1. Given information.

given,

∑k=1∞ (3k−1)−2/3

02

Step 2. The objective is to explain why the integral test is used to determine the convergence or divergence of the series and use the test to determine the convergence or divergence of the series. 

Consider function f(x)=(3x−1)−23.

The function f(x)=(3x−1)−23is continuous, decreasing, with positive terms. Therefore, all the conditions of the integral test are fulfilled. So, the integral test is applicable.

03

Step 3.  Consider the integral ∫x=1∞ f(x)dx=∫x=1∞ (3x−1)−23dx.

Therefore,

∫x=1∞ f(x)dx=limk→∞ ∫x=1k (3x−1)−23dx=13limk→∞ ∫u=23k−1 u−2/3du(Put3x−1=u⇒3dx=du=limk→∞ u1/323k−1(Integrating)=limk→∞ (3k−1)1/3−21/3(Substitution)=∞

04

Step 4. Thus, the value of the integral is ∫x=1∞ (3x−1)−23dx=∞

The integral converges. Therefore, the series ∑k=1∞ (3k−1)−23is divergent.

Hence, by integral test, the series∑k=1∞ (3k−1)−23is divergent.

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Most popular questions from this chapter

Which p-series converge and which diverge?

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 ak→0, then ∑k=1∞akconverges.

(b) True or False: If ∑k=1∞akconverges, then ak→0.

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

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

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

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

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

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

Iflimk→∞akbk=∞and∑k=1∞_____Convergesthen∑k=1∞_____(converges/diverges).

Express each of the repeating decimals in Exercises 71–78 as a geometric series and as the quotient of two integers reduced to lowest terms.

2.2131313...

In Exercises 48–51 find all values of p so that the series converges.

∑k=1∞1(c+k)p,wherec>0
See all solutions

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