To determine whether the given series converges or diverges, we need to analyze:

$\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^7 + 2}}$

Step 1: Simplifying the General Term

Let’s examine the general term: $a_n = \frac{n}{\sqrt{n^7 + 2}}$

For large $n$, the term $n^7$ dominates over the constant $2$ inside the square root. So we can approximate the denominator for large $n$ as follows: $\sqrt{n^7 + 2} \approx \sqrt{n^7} = n^{7/2}$

Thus, for large $n$, we can approximate $a_n$ by: $a_n \approx \frac{n}{n^{7/2}} = n^{1 - 7/2} = n^{-5/2}$

Step 2: Applying the Comparison Test

The series $\sum \frac{1}{n^{5/2}}$ is a p-series with $p = 5/2$, which converges because $p > 1$.

Since $a_n \approx \frac{1}{n^{5/2}}$ for large $n$, we can use the Limit Comparison Test with the series $\sum \frac{1}{n^{5/2}}$.

Step 3: Limit Comparison Test

We compute the limit: $\lim_{n \to \infty} \frac{a_n}{\frac{1}{n^{5/2}}} = \lim_{n \to \infty} \frac{\frac{n}{\sqrt{n^7 + 2}}}{\frac{1}{n^{5/2}}} = \lim_{n \to \infty} \frac{n \cdot n^{5/2}}{\sqrt{n^7 + 2}} = \lim_{n \to \infty} \frac{n^{7/2}}{\sqrt{n^7 + 2}}$

Since $\sqrt{n^7 + 2} \approx n^{7/2}$ as $n \to \infty$, this limit becomes: $\lim_{n \to \infty} \frac{n^{7/2}}{n^{7/2}} = 1$

Since the limit is a positive finite number (1), the Limit Comparison Test tells us that $\sum a_n$ converges if and only if $\sum \frac{1}{n^{5/2}}$ converges.

Conclusion

The series $\sum \frac{1}{n^{5/2}}$ converges, so the original series $\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^7 + 2}}$ also converges.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How does the Limit Comparison Test differ from the Direct Comparison Test?
2. Why does the p-series test determine convergence based on the exponent $p$?
3. What would happen if the power of $n$ in the denominator was different, say $n^3$ instead of $n^7$?
4. Can the Root Test be applied to this series, and would it yield the same result?
5. How would you determine convergence if the series had an added oscillating factor, such as $(-1)^n$ in the numerator?

Tip: When using the Limit Comparison Test, ensure the terms you compare approach zero at the same rate for the test to be effective.