Consider


∑ [(3k+5)/(k²-2k)]ᵖ, for each p ∈ ℝ.
k=3

Show this series { converges if p > 1
{ diverges if p ≤ 1

Hint: Determine the known series whose terms past the second give an approximate match for the terms of this series. This series is suitable (almost) for using the comparison test. Separate comparisons with it, or closely related series, are needed to establish convergence or divergence of the series according to p. You will need to establish inequalities, based on approximations (as below), to apply the comparison tests.

k² - 2k ≥ k²/2 for k ≥ 4 3k + 5 ≤ 8k

k² > k² - 2k k < 3k + 5

Those approximations are supposed to look like this:

k² - 2k ≥ k²/2 for k ≥ 4

3k + 5 ≤ 8k

k² > k² - 2k

k < 3k + 5

To determine if the given series converges or diverges based on the value of p, we can use the comparison test. The comparison test states that if 0 ≤ a_k ≤ b_k for all k, and the series Σ b_k converges, then the series Σ a_k also converges. On the other hand, if the series Σ b_k diverges, then the series Σ a_k also diverges.

Let's start by finding a known series whose terms past the second give an approximate match for the terms of the given series. In this case, we can consider the series Σ 1/k^q, where q is a positive real number. This series is well-known and its convergence depends on the value of q.

Now, let's determine the inequalities based on approximations to apply the comparison test. We have:

k^2 - 2k ≥ k^2/2 for k ≥ 4 (Inequality 1)

This inequality follows from the fact that for k ≥ 4, subtracting 2k from k^2 results in at least k^2/2.

Similarly, we have:

3k + 5 ≤ 8k (Inequality 2)

This inequality follows from the fact that adding 5 to 3k results in less than or equal to 8k.

Now, let's rewrite the terms of the given series using the inequalities:

[(3k+5)/(k^2-2k)]^p ≥ [(8k)/(k^2/2)]^p
= (16k^3p)/(k^2)^p
= 16k^(3p-2p)
= 16k^p

Now, we can compare the rewritten terms to the terms of the series Σ 1/k^q, where q is a positive real number. We want to determine the value of q that makes the inequality:

16k^p ≥ 1/k^q

Hold true for all positive integers k ≥ 4. In other words, we want to find the smallest q such that the above inequality holds true.

From the inequality, we can see that k^p ≤ 1/(16k^q). Rearranging the terms, we have:

k^(p+q) ≤ 1/16

Taking the logarithm of both sides, we get:

(p+q) log(k) ≤ log(1/16)

Since we want the inequality to hold for all k ≥ 4, we can choose the smallest p+q such that the above inequality holds true.

Therefore, we have:

p+q > -4

Since p and q are positive real numbers, we conclude that p > -4.

Now, let's consider the two cases separately:

Case 1: p > 1
In this case, the series Σ 1/k^q converges for q < 1. Since we have shown that 16k^p ≥ 1/k^q for q < 1, it follows that the given series Σ [(3k+5)/(k^2-2k)]^p converges by the comparison test.

Case 2: p ≤ 1
In this case, the series Σ 1/k^q diverges for q ≥ 1. Since we have shown that 16k^p ≥ 1/k^q for q ≥ 1, it follows that the given series Σ [(3k+5)/(k^2-2k)]^p also diverges by the comparison test.

Therefore, the series Σ [(3k+5)/(k^2-2k)]^p converges if p > 1 and diverges if p ≤ 1.