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

Conditionally convergent

Absolutely convergent

Divergent

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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.

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.

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.

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.

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.'

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