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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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