Use the Alternating Series Test to determine if the following series is conditionally convergent, absolutely convergent, or divergent.
${\sum }_{k=1}^{\infty }\frac{\sin \frac{n\pi }{2}}{n^3}$

• A. Conditionally convergent
• B. Absolutely convergent
• C. Divergent

Answer: (B)

Verified step by step guidance

1. Step 1: Recall the Alternating Series Test (Leibniz Criterion). It states that an alternating series ∑(-1)^(n-1)a_n converges if: (1) The terms a_n are positive, (2) The terms a_n are decreasing, and (3) The limit of a_n as n approaches infinity is zero.
2. Step 2: Analyze the given series ∑k=1∞(sin(nπ/2)/n^3). Notice that the numerator sin(nπ/2) alternates between 0, 1, and -1 depending on the value of n. This suggests the series is not strictly alternating in sign, so we need to carefully examine its behavior.
3. Step 3: Check the absolute convergence of the series. To do this, consider the series ∑k=1∞|sin(nπ/2)/n^3|. Since |sin(nπ/2)| is bounded by 1, the series becomes ∑k=1∞(1/n^3), which is a p-series with p = 3. Recall that a p-series converges if p > 1.
4. Step 4: Conclude that the series ∑k=1∞|sin(nπ/2)/n^3| converges absolutely because the p-series ∑k=1∞(1/n^3) converges. Absolute convergence implies that the original series is absolutely convergent.
5. Step 5: Summarize the findings. Since the series converges absolutely, it is not conditionally convergent or divergent. The correct classification of the series is 'Absolutely convergent.'