We are tasked with computing the given limit to analyze the series' convergence using the limit comparison test.

The limit to compute:

$\lim_{n \to \infty} \frac{\frac{n^3 - 1}{n^8 - 6}}{\frac{1}{n^5}}$

This simplifies to: $\lim_{n \to \infty} \frac{n^3 - 1}{n^8 - 6} \cdot n^5$

Simplifying further:

1. Multiply $n^5$ into the numerator: $\frac{(n^3 - 1) n^5}{n^8 - 6} = \frac{n^{8} - n^5}{n^8 - 6}.$

2. Divide both numerator and denominator by $n^8$ to handle the $n \to \infty$ case: $\frac{\frac{n^8 - n^5}{n^8}}{\frac{n^8 - 6}{n^8}} = \frac{1 - \frac{1}{n^3}}{1 - \frac{6}{n^8}}.$

Taking the limit as $n \to \infty$:

• As $n \to \infty$, $\frac{1}{n^3} \to 0$ and $\frac{6}{n^8} \to 0$.
• Therefore, the limit becomes: $\frac{1 - 0}{1 - 0} = 1.$

Conclusion:

The limit is: $\boxed{1}$

Since the limit is finite and positive ($1 > 0$), the comparison test tells us that the convergence behavior of $\sum \frac{n^3 - 1}{n^8 - 6}$ matches that of $\sum \frac{1}{n^5}$.

Because $\sum \frac{1}{n^5}$ converges (it is a $p$-series with $p = 5 > 1$), the given series also converges.

Would you like further clarification or details? Here are five related questions to explore further:

1. How does the $p$-series test work for determining convergence?
2. What other comparison tests can be used to analyze series convergence?
3. Can the limit comparison test fail in any situation?
4. Why is dividing by the highest power of $n$ in limits effective?
5. What are some examples of divergent series for comparison?

Tip: Always simplify the expressions in a series or sequence to its leading terms when handling limits as $n \to \infty$.