Answer :
Answer
hope this helps :)
C pretty simple
Step-by-step explanation:
To determine the convergence of the series provided, we can use different convergence tests. Let's analyze each series:
1. ∑n=1∞(-1)^(n)/(4n+3)
This is an alternating series. By the Alternating Series Test, the series converges as the terms decrease towards zero as n approaches infinity. Therefore, the correct answer is B. The series converges, but is not absolutely convergent.
2. ∑n=1∞(-1)^(n)((n)^(.5)/(n+5)
This series involves the ratio of square root terms. By comparing the highest power terms in the numerator and denominator, we see that the series behaves like ∑n=1∞(-1)^(n)/(n^(1.5)). Since the power of n is greater than 1, the series diverges. Therefore, the correct answer is D. The series diverges.
3. ∑n=1∞(-2)^(n)/(n^(7))
This series involves exponential terms. As the absolute value of the terms does not approach zero, the series diverges. Therefore, the correct answer is D. The series diverges.
4. ∑n=1∞sin(5n)/(n^(2))
Since the sine function oscillates between -1 and 1, the terms in the series do not approach zero. Thus, the series diverges. Therefore, the correct answer is D. The series diverges.
5. ∑n=1∞((n+1)(2^(2)-1)^(n))/(2^(2n))
By simplifying the expression, we get ∑n=1∞((n+1)3^(n))/(4^n). Comparing the powers of n and simplifying further, we see that the series behaves like ∑n=1∞(n/4)^n. Since the power of n is 1, the series converges. However, it is not absolutely convergent. Therefore, the correct answer is C. The series converges, but is not absolutely convergent.