Answer:
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<P<=1 diverges and (ii) is satisfied. Hence Converges
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<P<=1 diverges and (ii) is satisfied. Hence Converges.
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
