Could you please circle the answers? Thank you! Assignment 3: Problem 5 Previous

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Could you please circle the answers? Thank you!

Assignment 3: Problem 5 Previous Problem Problem List Next Problem (1 point) Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) In(n)1 In(n) 1. For all n > 2 2. For all n > 1 > , and the series diverges, so by the Comparison Test, the series 11) diverges n? 2 2n3 , -6 nand the series n, converges, so by the Comparison Test, the series ! converges - In(n 1 In(n) 5 For all n>2.1 6. For all n > 2, T 15 and the series 1 converges, so by the Comparison Test, the series -converges EE converges, so by the Comparison Test, the series converges

Explanation / Answer

lnn/n > 1/n indeed for all n > 2, this is true.
And we know 1/n diverges
So, a greater series like ln(n)/n also diverges
So, yes, this is TRUE
C

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When we grpah lnx/x^2 and 1/x^1.5, we notice that as far as the eye can see, the inequality holds good
i.e ln(n)/n^2 < 1/n^1.5 is true
And we know 1/n^1.5 is convergent by p-series test
So, anything smaller like ln(n)/n^2 also CONVERGES
C

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Notice that n/(2-n^3) can also come out negative
But for comparison test, we need them to both be >= 0
So, comparison test cant be applied
I

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C

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I

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1/(n^2 - 5) is in no way smaller than 1/n^2
because n^2 - 5 in the denominator is smaller than n^2
So, 1/n^2-5 is actually larger
So,
I

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