To determine whether the series

$\sum_{n=1}^{\infty} \frac{1}{2n^3 + 8}$

is convergent or divergent, let's analyze its behavior using the comparison test or limit comparison test with a simpler, comparable series.

Step-by-Step Solution

1. Identify the Dominant Term in the Denominator: For large values of $n$, the $2n^3$ term dominates the $8$ term in the denominator, so the expression $\frac{1}{2n^3 + 8}$ behaves similarly to $\frac{1}{2n^3}$ as $n \to \infty$.

2. Simplify the Comparison: We can compare $\frac{1}{2n^3 + 8}$ with the series $\frac{1}{2n^3}$, which simplifies to $\frac{1}{n^3}$ after factoring out the constant.

3. Determine Convergence of the Comparison Series: The series $\sum \frac{1}{n^3}$ is a p-series with $p = 3$. Since $p > 1$, this series is convergent.

4. Apply the Limit Comparison Test: We apply the limit comparison test between $\frac{1}{2n^3 + 8}$ and $\frac{1}{n^3}$:

   $\lim_{n \to \infty} \frac{\frac{1}{2n^3 + 8}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{n^3}{2n^3 + 8} = \lim_{n \to \infty} \frac{1}{2 + \frac{8}{n^3}} = \frac{1}{2}$

   Since this limit is a positive, finite number, both series will converge or diverge together.

5. Conclusion: Since $\sum \frac{1}{n^3}$ converges, by the limit comparison test, $\sum_{n=1}^{\infty} \frac{1}{2n^3 + 8}$ also converges.

Answer

The series $\sum_{n=1}^{\infty} \frac{1}{2n^3 + 8}$ is convergent.

Would you like more details on this process or have any specific questions?

Here are five related questions to explore further:

1. What is the p-series test, and how can it help determine convergence?
2. How does the comparison test work, and when is it most useful?
3. Can you apply the ratio test to determine convergence for this series?
4. What changes in the series form would lead to divergence instead?
5. How does the behavior of the dominant term influence the convergence of a series?

Tip: When using the comparison or limit comparison tests, always look for the dominant term in the denominator (or numerator) to find a comparable series.