Determine the following about the series. Indicate the test that was used and justify your answer.

Sigma (lower index n = 1; upper index infinity) [sin((2n-1)pi/2)]/n

A. The series diverges
B. The series converges conditionally.
C. The series converges absolutely.
D. It cannot be determined.

since sin(nth odd multiple of pi/2) = -(-1)^n the sequence is

1 - 1/2 + 1/3 - 1/4 +...

The alternating harmonic series converges to ln(2)

To determine the convergence or divergence of the series:

1. Start by applying the Limit Comparison Test. Let's compare the given series with the harmonic series, which is given by Σ(1/n).

2. Write out the limit comparison setup:
lim (as n approaches infinity) [sin((2n-1)π/2)]/n ÷ (1/n)

3. Simplify the expression inside the limit:
lim (as n approaches infinity) [sin((2n-1)π/2)]

4. Since the argument of the sine function, (2n-1)π/2, cycles between π/2 and -π/2 as n increases, the sine function oscillates indefinitely and does not approach a specific value. Therefore, the limit does not exist.

5. Based on the limit from step 4, we cannot conclude whether the series converges or diverges. However, we can determine if it converges absolutely or conditionally by examining the behavior of the absolute series.

6. Consider the absolute series:
Σ |[sin((2n-1)π/2)]/n|

7. The absolute value of the terms within the series is |sin((2n-1)π/2)|/n, which simplifies to 1/n.

8. The absolute series can be compared to the harmonic series, Σ(1/n). However, we already know that the harmonic series diverges.

9. Since the harmonic series diverges, and the absolute series is larger or equal to it, we can conclude that the absolute series also diverges.

10. Therefore, the given series, Σ[sin((2n-1)π/2)]/n, diverges.

Answer: A. The series diverges.