( .) 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.) √n+1 1. For all n > 2, n by the Comparison Test, the series > sin² (n) n² 2. For all n > 1, so by the Comparison Test, the series 1 n ln(n) by the Comparison Test, the series > 1 n² - 2 so by the Comparison Test, the series > n 5. For all n > 1, 3-n³ so by the Comparison Test, the series 3. For all n > 1, 4. For all n > 2, 6. For all n > 2, 1 n √n +1 n 1 n n³ - 1 so by the Comparison Test, the series > 1 n² sin²(n) n² 1 7² and the series nln(n) 1 n² 2 n² 2 <, and the series 2 = diverges, so n n diverges. , and the series Σ diverges. and the series Σ 1 n² - 2 1 1 1 1 Σ; diverges, so n n 3-n³ converges. n n³ - 1 converges. and the series Σ/1/2 n² converges. and the series 2 n² converges. converges, converges, converges, 1 n² converges,
( .) 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.) √n+1 1. For all n > 2, n by the Comparison Test, the series > sin² (n) n² 2. For all n > 1, so by the Comparison Test, the series 1 n ln(n) by the Comparison Test, the series > 1 n² - 2 so by the Comparison Test, the series > n 5. For all n > 1, 3-n³ so by the Comparison Test, the series 3. For all n > 1, 4. For all n > 2, 6. For all n > 2, 1 n √n +1 n 1 n n³ - 1 so by the Comparison Test, the series > 1 n² sin²(n) n² 1 7² and the series nln(n) 1 n² 2 n² 2 <, and the series 2 = diverges, so n n diverges. , and the series Σ diverges. and the series Σ 1 n² - 2 1 1 1 1 Σ; diverges, so n n 3-n³ converges. n n³ - 1 converges. and the series Σ/1/2 n² converges. and the series 2 n² converges. converges, converges, converges, 1 n² converges,
Chapter6: Exponential And Logarithmic Functions
Section6.2: Graphs Of Exponential Functions
Problem 52SE: Prove the conjecture made in the previous exercise.
Related questions
Question
![( ..) 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 1.)
1. For all n > 2,
2. For all n > 1,
by the Comparison Test, the series >
sin² (n)
n²
3. For all n > 1,
√n +1 1
>
so by the Comparison Test, the series
1
n ln(n)
n
4. For all n > 2,
by the Comparison Test, the series >
6. For all n > 2,
n
5. For all n > 1,
3-n³
so by the Comparison Test, the series
1
n² - 2
so by the Comparison Test, the series >
n
<
n³ 1
so by the Comparison Test, the series
-
n
n+1
n
1
[१२]
"
2
n
"
1
sin² (n)
n²
and the series 2
n²
1
and the series Σ diverges, so
n ln(n)
1
"
n²
1
1
diverges.
and the series
1
n² - 2
1
n²
diverges.
n
3- n³
2
1
Στ converges,
converges.
and the series
and the series Σ converges,
n
n³ - 1
n
1
Σdiverges, so
converges.
Σ
converges.
1
n²
1
n²
converges,
and the series 2 Σ;
n²
converges.
converges,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa38b1750-4836-4d47-bfc9-b05e81f0daae%2F59f0c008-e1da-475d-9b46-d2fe7e826b8b%2Fk4h4vt2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:( ..) 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 1.)
1. For all n > 2,
2. For all n > 1,
by the Comparison Test, the series >
sin² (n)
n²
3. For all n > 1,
√n +1 1
>
so by the Comparison Test, the series
1
n ln(n)
n
4. For all n > 2,
by the Comparison Test, the series >
6. For all n > 2,
n
5. For all n > 1,
3-n³
so by the Comparison Test, the series
1
n² - 2
so by the Comparison Test, the series >
n
<
n³ 1
so by the Comparison Test, the series
-
n
n+1
n
1
[१२]
"
2
n
"
1
sin² (n)
n²
and the series 2
n²
1
and the series Σ diverges, so
n ln(n)
1
"
n²
1
1
diverges.
and the series
1
n² - 2
1
n²
diverges.
n
3- n³
2
1
Στ converges,
converges.
and the series
and the series Σ converges,
n
n³ - 1
n
1
Σdiverges, so
converges.
Σ
converges.
1
n²
1
n²
converges,
and the series 2 Σ;
n²
converges.
converges,
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Follow-up Question
![( ..) 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 1.)
1. For all n > 2,
2. For all n > 1,
by the Comparison Test, the series >
sin² (n)
n²
3. For all n > 1,
√n +1 1
>
so by the Comparison Test, the series
1
n ln(n)
n
4. For all n > 2,
by the Comparison Test, the series >
6. For all n > 2,
n
5. For all n > 1,
3-n³
so by the Comparison Test, the series
1
n² - 2
so by the Comparison Test, the series >
n
<
n³ 1
so by the Comparison Test, the series
-
n
n+1
n
1
[१२]
"
2
n
"
1
sin² (n)
n²
and the series 2
n²
1
and the series Σ diverges, so
n ln(n)
1
"
n²
1
1
diverges.
and the series
1
n² - 2
1
n²
diverges.
n
3- n³
2
1
Στ converges,
converges.
and the series
and the series Σ converges,
n
n³ - 1
n
1
Σdiverges, so
converges.
Σ
converges.
1
n²
1
n²
converges,
and the series 2 Σ;
n²
converges.
converges,](https://content.bartleby.com/qna-images/question/a38b1750-4836-4d47-bfc9-b05e81f0daae/6f6fe79b-f39e-429c-b475-5c74c585599a/bx8kq09_processed.jpeg)
Transcribed Image Text:( ..) 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 1.)
1. For all n > 2,
2. For all n > 1,
by the Comparison Test, the series >
sin² (n)
n²
3. For all n > 1,
√n +1 1
>
so by the Comparison Test, the series
1
n ln(n)
n
4. For all n > 2,
by the Comparison Test, the series >
6. For all n > 2,
n
5. For all n > 1,
3-n³
so by the Comparison Test, the series
1
n² - 2
so by the Comparison Test, the series >
n
<
n³ 1
so by the Comparison Test, the series
-
n
n+1
n
1
[१२]
"
2
n
"
1
sin² (n)
n²
and the series 2
n²
1
and the series Σ diverges, so
n ln(n)
1
"
n²
1
1
diverges.
and the series
1
n² - 2
1
n²
diverges.
n
3- n³
2
1
Στ converges,
converges.
and the series
and the series Σ converges,
n
n³ - 1
n
1
Σdiverges, so
converges.
Σ
converges.
1
n²
1
n²
converges,
and the series 2 Σ;
n²
converges.
converges,
Solution
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