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Chapter 7: Q. 33 (page 639)

In Exercises 29–34 use the ratio test to analyze whether the given series converges or diverges. If the ratio test is inconclusive, use a different test to analyze the series.

∑k=1∞2·4·6⋯(2k)1·3·5⋯(2k-1)

Short Answer

Expert verified

The series diverges.

Step by step solution

01

Step 1. Given information.

The given series is∑k=1∞2·4·6⋯(2k)1·3·5⋯(2k-1).

02

Step 2. Ratio Test.

According to the series,

ak+1ak=2·4·6…(2k)(2k+2)1·3·5·…·(2k-1)(2k+1)2·4·6⋯(2k)1·3·5·⋯(2k-1)=2k+22k+1limk→∞ak+1ak=limk→∞2k+22k+1=limk→∞k2+2kk2+1k=1

Therefore, the test is inconclusive.

03

Step 3. Divergence Test.

Now,

ak=2·4·6·…(2k)1·3·5·…(2k-1)limk→∞ak≠0Here,akdoes not converge to0.

Hence, the series is divergent.

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Most popular questions from this chapter

Determine whether the series ∑k=0∞πe3kconverges or diverges. Give the sum of the convergent series.

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.

Given thata0=-3,a1=5,a2=-4,a3=2and∑akk=2∞=7, find the value ofrole="math" localid="1648828803227" ∑akk=3∞.

Determine whether the series ∑n=4∞411nconverges or diverges. Give the sum of the convergent series.

Determine whether the series∑k=0∞5k+1-6k converges or diverges. Give the sum of the convergent series.
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