To analyze whether the given series

$\sum_{n=1}^{\infty} \frac{\sqrt{3n^3 + 15n + 1}}{4n^9 + 4n}$

is convergent or divergent, we need to choose an appropriate convergence test. Here’s a structured approach to evaluating this series:

1. Identify the Dominant Terms:

   • The dominant term in the numerator, $\sqrt{3n^3 + 15n + 1}$, is $\sqrt{3n^3} = n^{3/2}$.
   • The dominant term in the denominator, $4n^9 + 4n$, is $4n^9$.

   This simplifies the terms to $\frac{n^{3/2}}{n^9} = \frac{1}{n^{15/2}}$ for large $n$.

2. Choose a Convergence Test:

   • Since the series resembles a $p$-series, we can use the Limit Comparison Test with $b_n = \frac{1}{n^{15/2}}$.

3. Applying the Limit Comparison Test:

   • Calculate $\lim_{n \to \infty} \frac{a_n}{b_n}$, where $a_n = \frac{\sqrt{3n^3 + 15n + 1}}{4n^9 + 4n}$ and $b_n = \frac{1}{n^{15/2}}$.
   • If this limit is finite and positive, then the convergence behavior of $\sum a_n$ will match that of $\sum b_n$, which converges for $p = 15/2 > 1$.

Let me know if you would like to see the calculations for the limit or further breakdown of the steps.

Follow-Up Questions

1. What are the steps in finding the dominant term in a series?
2. How does the Limit Comparison Test work in identifying convergence?
3. Why does the $p$-series converge for $p > 1$?
4. Can this series be tested with the Integral Test instead?
5. How can you apply the Ratio Test to identify the behavior of a series?

Tip

The Limit Comparison Test is particularly useful when the terms of the series have similar forms, especially for complex expressions involving roots and powers.