To determine whether the series

$\sum_{n=1}^{\infty} \frac{\sqrt{n^2 + 2n + 2}}{4n^3 - 2n^2 + n}$

converges or diverges, let's analyze the behavior of the terms $\frac{\sqrt{n^2 + 2n + 2}}{4n^3 - 2n^2 + n}$ as $n \to \infty$. For large $n$, the highest-degree terms in both the numerator and denominator dominate, so let's approximate:

Step 1: Simplify the Expression for Large $n$

1. Numerator Analysis: $\sqrt{n^2 + 2n + 2} \approx \sqrt{n^2} = n$ as $n$ grows large, because $n^2$ is the leading term.

2. Denominator Analysis: $4n^3 - 2n^2 + n \approx 4n^3$ for large $n$, since $4n^3$ is the leading term.

Step 2: Approximate the Term

Using the approximations above, we find that $\frac{\sqrt{n^2 + 2n + 2}}{4n^3 - 2n^2 + n} \approx \frac{n}{4n^3} = \frac{1}{4n^2}.$

Step 3: Compare with a Known Convergent Series

Now, we can compare this with the series $\sum \frac{1}{4n^2}$, which simplifies to a constant multiple of the p-series $\sum \frac{1}{n^2}$ (with $p = 2$). Since $p > 1$, the p-series converges.

Conclusion

Since $\frac{\sqrt{n^2 + 2n + 2}}{4n^3 - 2n^2 + n}$ behaves similarly to $\frac{1}{4n^2}$ for large $n$, and since $\sum \frac{1}{n^2}$ converges, we conclude that the original series also converges by the Limit Comparison Test.

Would you like further details on any specific part of this process?

Here are five questions that build on this topic:

1. What is the Limit Comparison Test, and how is it applied to determine convergence?
2. How would we handle this problem if the degree of the numerator and denominator were different?
3. What are p-series, and why are they important in convergence tests?
4. How does the behavior of a series term's highest degree influence convergence?
5. Could the Direct Comparison Test be applied to this series instead?

Tip: When determining convergence, always look at the dominant terms for large $n$, as they often dictate the series' behavior.