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Use any convergence tests to determine whether the series converge absolutely, converge conditionally, or diverge. Explain why the series meets the hypotheses of the test you select.

k=1-1kk2+3k+13k+1

Short Answer

Expert verified

The series converges conditionally.

Step by step solution

01

Step 1. Given information.

Consider the given question,

k=1-1kk2+3k+13k+1

02

Step 2. Use the alternating series test.

Using the alternating series test, assume akbe the sequence of positive numbers.

If ak+1<akfor every k1and limkak=0

Then the alternating series k=1-1k+1akand k=1-1kakboth converge.

Consider the series k=1-1kak=k=1-1kk2+3k+13k+1.

The sequence ak=k2+3k+13k+1is of positive terms.

Here, clearly the series is decreasing asak+1<ak.

03

Step 3. Find the value of limit.

The value of limkakis given below,

limkak=limkk2+3k+13k+1=0

Since all the conditions satisfied. Then, by alternating series test, the series is convergent.

As,

k=1-1kk2+3k+13k+1=k=1k2+3k+13k+1=k=1k2+3k+13k+1

04

Step 4. Consider the dominant term of numerator and denominator.

The terms of the series k=1k2+3k+13k+1 are positive.

The series k=1bkfor the series k=1k2+3k+13k+1is given by,

k=1bk=k=1k23k=k=11k13

05

Step 5. Find the ratio of the limit.

The ratio limkakbkis given by,

limkakbk=limkk2+3k+13k+11k13=limkk13×k2+3k+13k+1=limkk13×k23×1+3k+1k23k+1=limk1+3k+1k231+1k=1

06

Step 6. Find the value of the limit which is non-zero finite number.

The value of limkakbk=1; which is non-zero finite number.

The series k=1bk=k=11k13 is divergent by p-series test.

Thus, the series k=1akis also divergent.

Hence, the series k=1k2+3k+13k+1is divergent.

Therefore, the given series is convergent conditionally.

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Most popular questions from this chapter

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.

Which p-series converge and which diverge?

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.

0.6345345...

For each series in Exercises 44–47, do each of the following:

(a) Use the integral test to show that the series converges.

(b) Use the 10th term in the sequence of partial sums to approximate the sum of the series.

(c) Use Theorem 7.31 to find a bound on the tenth remainder,R10.

(d) Use your answers from parts (b) and (c) to find an interval containing the sum of the series.

(e) Find the smallest value of n so thatRn10-6

k=11k2

Prove Theorem 7.24 (a). That is, show that if c is a real number andk=1ak is a convergent series, then k=1cak=ck=1ak.

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