We aim to compute the limit of the sequence $a_n = \frac{3n + 1}{2n^2 + n}$ as $n \to \infty$.

Step-by-Step Solution

1. Expression Analysis: The sequence is: $a_n = \frac{3n + 1}{2n^2 + n}.$

2. Simplify the Fraction: Divide the numerator and the denominator by the highest power of $n$ in the denominator, which is $n^2$: $a_n = \frac{\frac{3n}{n^2} + \frac{1}{n^2}}{\frac{2n^2}{n^2} + \frac{n}{n^2}} = \frac{\frac{3}{n} + \frac{1}{n^2}}{2 + \frac{1}{n}}.$

3. Take the Limit as $n \to \infty$: As $n \to \infty$, the terms $\frac{3}{n}$, $\frac{1}{n^2}$, and $\frac{1}{n}$ approach $0$. Hence, the sequence simplifies to: $\lim_{n \to \infty} a_n = \frac{0 + 0}{2 + 0} = \frac{0}{2} = 0.$

Final Answer:

$\boxed{0}$

Do you want further details or have any questions?
Here are 5 related questions to deepen your understanding:

1. What happens to the sequence if the numerator has a higher degree than the denominator?
2. How do you identify the dominant terms in the numerator and denominator of a sequence?
3. Can you explain why the smaller-order terms become negligible as $n \to \infty$?
4. What is the limit of $b_n = \frac{2n^2 + 3}{4n^2 - n}$ as $n \to \infty$?
5. What is the significance of limits in analyzing the behavior of sequences?

Tip: When finding the limit of rational functions in sequences, always simplify by dividing through by the highest power of $n$ in the denominator!