Determine whether the series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3 - 4}$ converges or diverges.

1. Problem Description

Determine whether the series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3 - 4}$ converges or diverges.

2. Solution Steps

To determine whether the series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3-4}$ converges or diverges, we can use the limit comparison test. We compare it to the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a convergent p-series with $p=2>1$.

Let $a_n = \frac{3n+1}{n^3-4}$ and $b_n = \frac{1}{n^2}$.

First, note that for $n \ge 2$, $n^3 - 4 > 0$, so $a_n$ is positive. Also, $b_n$ is positive for all $n$.

Now we compute the limit:

$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{3n+1}{n^3-4}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{(3n+1)n^2}{n^3-4} = \lim_{n \to \infty} \frac{3n^3+n^2}{n^3-4}$$

Divide both numerator and denominator by $n^3$:

$$\lim_{n \to \infty} \frac{3+\frac{1}{n}}{1-\frac{4}{n^3}} = \frac{3+0}{1-0} = 3$$

Since $0 < 3 < \infty$, the limit comparison test tells us that $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ either both converge or both diverge.

Since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent p-series with $p=2>1$, the series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3-4}$ also converges.

3. Final Answer

Converges