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.)

1. For all n>2, ln(n)/n>1n, and the series ∑1/n diverges, so by the Comparison Test, the series ∑ln/(n)n diverges.
2. For all n>1, arctan(n)/n3<π2n3, and the series π/2∑1/n3 converges, so by the Comparison Test, the series ∑arctan(n)/n3 converges.
3. For all n>1, n/(2−n3)<1n2, and the series ∑1/n2 converges, so by the Comparison Test, the series ∑n/(2−n3) converges.
4. For all n>1, ln(n)/n2<1/n1.5, and the series ∑1/n1.5 converges, so by the Comparison Test, the series ∑ln(n)/n2 converges.
5. For all n>1, 1/nln(n)<2/n, and the series 2∑1/n diverges, so by the Comparison Test, the series ∑1/nln(n) diverges.
6. For all n>2, 1/(n2−7)<1/n2, and the series ∑1/n2 converges, so by the Comparison Test, the series ∑1/(n2−7) converges.

1. C

2. C

3. C

4. C

5. I

6. I

Step-by-step explanation:.

Comparison theorem

Comparison test states that

Let, 0 ≤ an ≤ bn

For all n

Then, if

Σbn converges then, Σan converge

Σan diverges then, Σbn diverge

I.e if the small series diverges, then, the big series diverges and if the big series converges then, the small series converges.

P series says that,

Σ 1/n^p

If p≤1 then the series diverges

If p>1 the series converges

1. Given that,

For all n>2, ln(n)/n > 1 / n, and the series ∑1/n diverges, so by the Comparison Test, the series ∑ln(n)/n diverges.

Solution

In(n) / n > 1 / n. This is correct for n>2

Also, ∑1/n diverges. This is also true from P series, since p = 1, then, 1/n diverges

Now, it is shown that, the smaller series diverges I.e 1/n, then the bigger series In(n) / n also diverges

So, this is true.

C.

2. For all n>1, arctan(n)/n³ < π/2n³, and the series π/2∑1/n³ converges, so by the Comparison Test, the series ∑arctan(n)/n³ converges.

Solution

arctan(n)/n³ < π/2n³, let check of this is always correct for all n>1. This is always true since, arctan(n) is always less than π/2 if n>1

arctan(2) = 1.107 < π/2

Arctan(999) = 1.5697 < π/2

Then, π/2∑1/n³ converges. This is correct, because from p series, if p>1, the series converges since p = 3.

So, since the bigger series π/2n³ converges, then the smaller series (arctan(n)/n³) will converge, so with comparison test we conclude that it will converge.

So, it is correct

C

3. For all n>1, n/(2−n3)<1n2, and the series ∑1/n2 converges, so by the Comparison Test, the series ∑n/(2−n3) converges.

Solution

n/(2-n³) < 1/n³ for n > 1

When n = 2,3,4

This is always true because as n goes large the left hand side becomes negative and it will be less than the right hand side

Then, ∑1/n³ converges. This is correct, because from p series, if p>1, the series converges since p = 3..

Conclusion

So, since the bigger series 1/n³ converges, then the smaller series n/(2-n³) will converge, so with comparison test we conclude that it will converge.

So, it is correct

C

4. For all n>1, ln(n)/n² < 1/n1.5, and the series ∑1/n1.5 converges, so by the Comparison Test, the series ∑ln(n)/n2 converges.

Solution

ln(n)/n² < 1/n^1.5 for n>1

In(n) is greater than 1

Let try 3 and a big number 99

n=3

In3/3²< 1/3^1.5

0.1221 < 0.192 true

n=99

In99/99² < 1/99^1.5

4.69 × 10^-4 < 1.03 × 10^-3 true

So, the series comparison is true

So,

Then, ∑1/n^1.5 converges. This is correct, because from p series, if p>1, the series converges since p = 1.5

So, since the bigger series 1/n^1.5 converges, then the smaller series (In(n) / n²) will converge, so with comparison test we conclude that it will converge.

So, it is correct

C

5. For all n>1, 1/nln(n)<2/n, and the series 2∑1/n diverges, so by the Comparison Test, the series ∑1/nln(n) diverges.

Solution

1/nln(n) < 2 / n. For n>1

This is true

So, 2∑1/n diverges. This is correct, because from p series, if p≤1, the series diverges since p = 1

Now, the bigger series diverges, then, this doesn't implies that the smaller series will diverge, so, the comparison theorem doesn't accommodate this, so from comparison theorem it cannot concluded maybe smaller series diverges or not

So, the conclusions is that it is incorrect since one of the argument is incorrect.

I.

6. For all n>2, 1/(n2−7)<1/n2, and the series ∑1/n² converges, so by the Comparison Test, the series ∑1/(n2−7) converges.

Solution

1/(n²−7) < 1/n² for n>7

This is wrong n²-7 < 7

So, since one of the argument is wrong, then, we conclude that it is incorrect.

Though the other processes are correct. But since the first argument is wrong, then, it is not a valid argument

I.

i don't know i don't know

we know that

1) the inscribed angle measures half of the arc it comprises.

so

m∠dgf= > expression

2) the opposite angles in an inscribed quadrilateral are supplementary angles

so

m∠dgf=

m∠dgf=°

substitute in the expression to find the value of arc ef

therefore

the answer is

the measure of the arc ef is