We need to determine if the series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3-4}$ converges or diverges.

1. Problem Description

We need to determine if the series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3-4}$ converges or diverges.

2. Solution Steps

We can use the limit comparison test. Let $a_n = \frac{3n+1}{n^3-4}$. We will compare it to $b_n = \frac{1}{n^2}$. Note that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges because it is a $p$-series with $p=2 > 1$.

We calculate the limit:

$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{3n+1}{n^3-4}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{n^2(3n+1)}{n^3-4} = \lim_{n\to\infty} \frac{3n^3+n^2}{n^3-4} = \lim_{n\to\infty} \frac{3+\frac{1}{n}}{1-\frac{4}{n^3}} = \frac{3+0}{1-0} = 3$

Since the limit is a positive finite number (3), the series $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ either both converge or both diverge. Since $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, then $\sum_{n=1}^{\infty} \frac{3n+1}{n^3-4}$ also converges. Note that for $n=1$, the term is $\frac{4}{-3}$, for $n=2$, the term is $\frac{7}{4}$.

So, the original series starts from $n=3$ since $n^3-4>0$ when $n> \sqrt[3]{4}$.

Since we only are concerned with convergence, the series converges even if some initial terms may be undefined.

3. Final Answer

The series converges.