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Answers must be correct. Or else it will be flagged. All sub-parts need to be answered with step by step process showing all work and reasoning.

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Don't provide the exact answers that have been provided here already. Find the error and provide the right answers.

1.A.

1.B.

1.C.

At least one of the answers above is NOT correct. 1 point) We will determine whether the series 6" is convergent or divergent using the comparison test. Note that the series has positive terms, which is a requirement for applying the comparison test. First we must find an appropriate series b, for comparison (this series must also have growing parts of the numerator and denominator) Which of the following is true? OA.anbn for all n2 asonable choice is bn (eliminate everything but the fastest- B. an > b, for all n 2 1 C. none of the above The series ? bn is a divergent enter "convergent" or "divergent") series with ratio r-6 and hence it is divergent Finally, by the Comparison Test we conclude that the series ? an is "divergent") divergent (enter "convergent" or

Explanation / Answer

Starting from first :
Easiest to go with is...
bn = 6^n / 1^n or just 6N
So, bn = 6^n ----> ANS

Now, clearly
6^n / (1^n - 7n^2) > 6^n/1^n

an > bn ----> ANS

But 6^n/1^n diverges

So, by comparison,
an = 6^n/(1^n - 7n^2) also diverges

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1B)
WE can go with

bn = bn = 4/n^(1/3) ----> ANS

Clearly
(lnn + 4)/n^(1/3) > 4/n^(1/3)

So,
an > bn -----> ANS

bn = divergent by p-series test

So, And thus by comparison test
an DIVERGES ---> ANS

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1C)
bn = n^(3/4) / n^2
i.e we can go with bn = 1/n^(5/4) ----> ANS

The limit is 1/2 ---> ANS

bn is 1/n^(5/4) is convergent

So, an is also CONVERGENT -----> ANS

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