Select the FIRST correct reason why the given series converges. A. Convergent geometric series B. Convergent p series C. Comparison (or Limit Comparison) with a geometric or p series D. Converges by alternating series test 1. n=1 2. Σ 3. n=1 ∞ n=1 ∞ +√n n4-2 (-1)" In(en) n5 cos(NT) 5(5)" 102n √n 4. (-1)". n+2 5. 6. n=1 ∞ n=1 n=1 (-1)" 5n+2 sin² (7n) n2 NOTE: the version of the alternating series test provided in section 11.5 of Stewart is not general enough to solve this problem. You will need the following version: If the series ✗(-1)^-16 m satisfies the conditions n=1 (i) there is an index N such that 0 < bn+1 ≤ bn for all n > N (ii) lim bn = 0 n→∞ then the series converges.

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Select the FIRST correct reason why the given series converges.
A. Convergent geometric series
B. Convergent p series
C. Comparison (or Limit Comparison) with a geometric or p series
D. Converges by alternating series test
1.
n=1
2. Σ
3.
n=1
∞
n=1
∞
+√n
n4-2
(-1)" In(en)
n5 cos(NT)
5(5)"
102n
√n
4. (-1)". n+2
5.
6.
n=1
∞
n=1
n=1
(-1)"
5n+2
sin² (7n)
n2
NOTE: the version of the alternating series test provided in section 11.5 of Stewart is not general enough to solve this problem. You
will need the following version:
If the series ✗(-1)^-16 m satisfies the conditions
n=1
(i) there is an index N such that 0 < bn+1 ≤ bn for all n > N
(ii) lim bn = 0
n→∞
then the series converges.
Transcribed Image Text:Select the FIRST correct reason why the given series converges. A. Convergent geometric series B. Convergent p series C. Comparison (or Limit Comparison) with a geometric or p series D. Converges by alternating series test 1. n=1 2. Σ 3. n=1 ∞ n=1 ∞ +√n n4-2 (-1)" In(en) n5 cos(NT) 5(5)" 102n √n 4. (-1)". n+2 5. 6. n=1 ∞ n=1 n=1 (-1)" 5n+2 sin² (7n) n2 NOTE: the version of the alternating series test provided in section 11.5 of Stewart is not general enough to solve this problem. You will need the following version: If the series ✗(-1)^-16 m satisfies the conditions n=1 (i) there is an index N such that 0 < bn+1 ≤ bn for all n > N (ii) lim bn = 0 n→∞ then the series converges.
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