Apply the divergence test and state what it tells you aboutthe series. ■5. (a)∑k=1k2 + k + 32k2 + 1

To apply the divergence test to the series ∑(k^2 + k + 3)/(2k^2 + 1), we need to check the behavior of the terms as k approaches infinity.

Step 1: Take the limit of the nth term as n approaches infinity. lim(n→∞) [(n^2 + n + 3)/(2n^2 + 1)]

Step 2: Simplify the expression inside the limit. lim(n→∞) [(1 + 1/n + 3/n^2)/(2 + 1/n^2)]

Step 3: Divide every term in the numerator and denominator by n^2. lim(n→∞) [(1/n^2 + 1/n + 3/n^2)/(2/n^2 + 1/n^4)]

Step 4: Take the limit of each term separately. lim(n→∞) [0 + 0 + 0]/[0 + 0] = 0/0

Step 5: Apply L'Hôpital's Rule to evaluate the limit. lim(n→∞) [(2/n^2 + 2/n^3)/(4/n^3)] = lim(n→∞) [(2n + 2)/(4n)] = 1/2

Step 6: Analyze the result. Since the limit of the nth term is not equal to zero, the series does not converge. Therefore, the divergence test tells us that the series ∑(k^2 + k + 3)/(2k^2 + 1) diverges.