For each of the following series, tell whether or not you can apply the 3-condition test (i. e. the alternating series test). Enter D if the series diverges by this test, C if the series converges by this test, and N if you cannot apply this test (even if you know how the series behaves by some other test).1. ∑n=2 to [infinity] ((-1) ^n (n^4+2^n)) / (n^3-1) 2. ∑n=1 to [infinity] (-1) ^n/n^53. ∑n=1 to [infinity] ((-1) ^ncos (n)) / (n^2) 4. ∑n=1 to [infinity] (((-1) ^n (n^3+1)) / (n^3+7) 5. ∑n=1 to [infinity] ((-1) ^n (n^10+1)) / e^n6. ∑n=1 to [infinity] (((-1) ^n (n^3+1)) / (n^4+1)

1. D 2. C 3. C 4. N 5. C 6. N

Step-by-step explanation:

The Alternating Series Test

∑ (-1) nBn converges when the following two conditions are met:

(i) lim Bn = 0 and (ii) {Bn} is (eventually) decreasing.

Note: The Alternating Series Test doesn't apply when either of the conditions is not met, and so never is a test for divergence. If the first condition isn't met, then the "n-th term test" will show divergence.

1. Bn = (n^4+2^n)) / (n^3-1)

lim (n^4+2^n)) / (n^3-1) = 0 but (ii) the function is increasing as n increases, Hence diverges

2. Bn = 1/n^5 lim Bn = 0. this is a converging p series as P>5 satisfies the condition for P series convergence P > 1 converges, 0
3. Bn = cos (n)) / (n^2); cos (n) has no limit since it dangles between 0 and 1 and limit is unique. leaving 1/n^2 to converge according to the condition for P series convergence P > 1 converges, 0
4. Bn = (n^3+1)) / (n^3+7); lim Bn as n tends to infinity = 0. (ii) can be confirmed by limit compression test

5. Bn = (n^10+1)) / e^n; lim Bn = 0 also Bn is a decreasing function since the denominator is an increasing function and (ii) is satisfied. Hence Converges.

6. Bn = (n^3+1)) / (n^4+1); lim Bn as n tends to infinity = 0. (ii) can be confirmed by limit compression test