Questions: Does the series sum from k=2 to infinity of (-1)^(k+1) * (k^4+2)/(k^5-1) converge absolutely, converge conditionally, or diverge? Choose the correct answer below and, if necessary, fill in the answer box to complete your choice. A. The series converges absolutely because of the Limit Comparison Test with sum from k=1 to infinity of 2/k^5. B. The series converges conditionally because of the Alternating Series Test and the Limit Comparison Test with sum from k=1 to infinity of 1/k. C. The series diverges because the limit used in the Divergence Test does not exist. D. The series converges absolutely because the limit used in the Divergence Test is

Solution Steps

To determine the convergence of the series \(\sum_{k=2}^{\infty}(-1)^{k+1} \frac{k^{4}+2}{k^{5}-1}\), we can use the following steps:

1. Absolute Convergence: Check if the series \(\sum_{k=2}^{\infty} \left|\frac{k^{4}+2}{k^{5}-1}\right|\) converges. This can be done using the Limit Comparison Test with a known convergent series, such as \(\sum_{k=1}^{\infty} \frac{1}{k}\).
2. Conditional Convergence: If the series does not converge absolutely, check for conditional convergence using the Alternating Series Test. This involves checking if the terms decrease in absolute value and if the limit of the terms as \(k\) approaches infinity is zero.
3. Divergence: If neither absolute nor conditional convergence is established, use the Divergence Test to check if the series diverges.

The given series is

\[ \sum_{k=2}^{\infty}(-1)^{k+1} \frac{k^{4}+2}{k^{5}-1} \]

This is an alternating series because of the factor \((-1)^{k+1}\).

To check for absolute convergence, we consider the series without the alternating sign:

\[ \sum_{k=2}^{\infty} \left| \frac{k^{4}+2}{k^{5}-1} \right| \]

We can use the Limit Comparison Test with the series \(\sum_{k=2}^{\infty} \frac{1}{k}\).

Consider the terms:

\[ a_k = \frac{k^{4}+2}{k^{5}-1}, \quad b_k = \frac{1}{k} \]

Calculate the limit:

\[ \lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\frac{k^{4}+2}{k^{5}-1}}{\frac{1}{k}} = \lim_{k \to \infty} \frac{k(k^{4}+2)}{k^{5}-1} = \lim_{k \to \infty} \frac{k^5 + 2k}{k^5 - 1} \]

Divide the numerator and the denominator by \(k^5\):

\[ = \lim_{k \to \infty} \frac{1 + \frac{2}{k^4}}{1 - \frac{1}{k^5}} = \frac{1 + 0}{1 - 0} = 1 \]

Since the limit is a positive finite number, the series \(\sum_{k=2}^{\infty} \frac{k^{4}+2}{k^{5}-1}\) behaves like \(\sum_{k=2}^{\infty} \frac{1}{k}\), which is a divergent harmonic series.

Since the series does not converge absolutely, we check for conditional convergence using the Alternating Series Test.

The Alternating Series Test requires:

1. \(b_k = \frac{k^{4}+2}{k^{5}-1}\) is positive.
2. \(b_k\) is decreasing.
3. \(\lim_{k \to \infty} b_k = 0\).

Check the limit:

\[ \lim_{k \to \infty} \frac{k^{4}+2}{k^{5}-1} = \lim_{k \to \infty} \frac{1 + \frac{2}{k^4}}{k - \frac{1}{k^5}} = 0 \]

The sequence \(b_k\) is positive and decreasing for \(k \geq 2\). Thus, the series converges conditionally.

Final Answer