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

We will use the Limit Comparison Test. We need to choose a comparison series. For large $n$, the term $\frac{3n+1}{n^3+4}$ behaves like $\frac{3n}{n^3} = \frac{3}{n^2}$. Therefore, we will compare the given series with $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which is a convergent $p$-series with $p=2 > 1$.

Let $a_n = \frac{3n+1}{n^3+4}$ and $b_n = \frac{1}{n^2}$. 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 the 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 the limit is 3, which is a positive finite number, 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 $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges (it is a $p$-series with $p=2>1$), the given series $\sum_{n=1}^{\infty} \frac{3n+1}{n^3+4}$ also converges.

3. Final Answer

The series converges.