Question

Determine whether the following series converge or diverge. $$\sum_{k=1}^{\infty} \frac{1}{(3 k+1)(3 k+4)}$$

Short answer

Question: Determine if the given series converges or diverges: $$\sum_{k=1}^{\infty}\frac{1}{(3 k+1)(3 k+4)}$$ Answer: The given series diverges.

Step-by-step solution

Set up the comparison series and inequality

Consider the harmonic series: $$\frac{1}{k}$$ Our goal is to compare this simpler series with the given series: $$\frac{1}{(3 k+1)(3 k+4)}$$ To do this, start by an inequality between the two series. The inequality should show that the given series is smaller term by term: $$\frac{1}{(3 k+1)(3 k+4)} < \frac{1}{k}$$

Solve for k

To solve for k, cross-multiply both sides of the inequality, and then simplify: $$k < (3k+1)(3k+4)$$ Expand and rearrange the inequality: $$0 < 9k^2 + 15k + 4 - k$$ Simplify: $$0 < 9k^2 + 14k + 4$$ Since the left side of the inequality is always positive (k starts from 1), the inequality is true for all k.

Apply the comparison test

Using the comparison test, if the harmonic series \(\frac{1}{k}\) diverges and the given series, $$\frac{1}{(3 k+1)(3 k+4)}$$ is smaller term by term, then the given series also diverges. The harmonic series is a well-known divergent series. Therefore, by the comparison test, the given series also diverges.