( ..) 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 so by the Comparison Test, the series 1 n ln(n) n 4. For all n > 2, by the Comparison Test, the series > 1 n² - 2 so 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 n n³ 1 so by the Comparison Test, the series - 1 1 > and the series Σ diverges, so < n √n +1 n 1 2 " n n2 sin² (n) n² and the series 2 " 1 n² n ln(n) 1 1 n² " diverges. and the series " 1 n² - 2 1 n² diverges. and the series n 3- n³ 2 converges. and the series 1 Στ converges, n - n³ - 1 converges. 1 Σdiverges, so n converges. 1 Σ²2 and the series 2 1 n² Σ; converges. converges, converges, 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 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 so by the Comparison Test, the series 1 n ln(n) n 4. For all n > 2, by the Comparison Test, the series > 1 n² - 2 so 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 n n³ 1 so by the Comparison Test, the series - 1 1 > and the series Σ diverges, so < n √n +1 n 1 2 " n n2 sin² (n) n² and the series 2 " 1 n² n ln(n) 1 1 n² " diverges. and the series " 1 n² - 2 1 n² diverges. and the series n 3- n³ 2 converges. and the series 1 Στ converges, n - n³ - 1 converges. 1 Σdiverges, so n converges. 1 Σ²2 and the series 2 1 n² Σ; converges. converges, converges, n² converges,
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![( ..) 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%2Fcb9ea953-4500-44af-863a-4cea4b6e0bbe%2F3mw3kue_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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