Question

(1 point) Each of the following statements is an attempt to show that a given series is convergent or divergent not 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.) I equation editorEquation Editor 1. For all n>2, nn3−2<2n2, and the series 2∑1n2 converges, so by the Comparison Test, the series ∑nn3−2 converges. equation editorEquation Editor 2. For all n>2, 1n2−1<1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑1n2−1 converges. I equation editorEquation Editor 3. For all n>2, n+1−−−−−√n>1n, and the series ∑1n diverges, so by the Comparison Test, the series ∑n+1−−−−−√n diverges. I equation editorEquation Editor 4. For all n>2, ln(n)n2>1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑ln(n)n2 converges. I equation editorEquation Editor 5. For all n>1, 1nln(n)<2n, and the series 2∑1n diverges, so by the Comparison Test, the series ∑1nln(n) diverges. I equation editorEquation Editor 6. For all n>1, n7−n3<1n2, and the series ∑1n2 converges, so by the Comparison Test, the series ∑n7−n3 converges.

Answer 1

Answer:

I.

CORRECT.

II.

CORRECT.

III.

CORRECT.

IV.

CORRECT.

V.

INCORRECT.

VI.

CORRECT

Step-by-step explanation:

To understand let us restate the comparison test in simple terms.

Comparison test :

Given     $\text{Series}_A$   and $\text{Series}_B$ such that    $\text{Series}_A < \text{Series}_B$ , then

1.   If    $\text{Series}_B$ converges then $\text{Series}_A$  converges as well.

2.  If  $\text{Series}_A$  diverges then $\text{Series}_B$  diverges as well.

Now to give you a more intuitive idea of what is going on, think about it like this.  When   the series on top converges it is like an "upper bound" for what you have on the bottom, therefore what you have on the bottom has to converge as well.

Similarly if what you have on the bottom explotes, then what you have on top will explote as well.

That's how I like to think about that intuitively.

Now, using those results let us examine the statements.

I.

$\frac{1}{n} < \frac{ln(n)}{n}$

Since the infinite sum of 1/n  diverges in fact the infinite sum of ln(n)/n does not converge.

Therefore, CORRECT.

II.

$\frac{\arctan(n)}{n^3} < \frac{\pi}{2}\frac{1}{n^3}$  

Since the infinite sum of    $\frac{\pi}{2}\frac{1}{n^3}$   is in fact convergent then  $\frac{\arctan(n)}{n^3}$ converges as well using the comparison theorem. Therefore

CORRECT.

III.

$\frac{n}{2-n^3} < \frac{1}{n^2}$

Once again   $1/n^2$  does converge so what you have on the bottom converges as well. Therefore

CORRECT.

IV.

$\frac{\ln(n)}{n^2} < \frac{1}{n^{1.5}}$

Once again   $\frac{1}{n^{1.5}}$  converges therefore since it is on top what is on the bottom converges as well. Therefore.

CORRECT.

V.

$\frac{\ln(n)}{n} < \frac{2}{n}$

Now the fact that  $\frac{2}{n}$    diverges does not necessarily imply that what you have on the bottom diverges. Therefore

INCORRECT.

VI.

That is correct as well since what you have on top converges therefore what you have on the bottom converges as well.