We can apply the limit comparison test to determine whether the series

$\sum_{n=2}^{\infty} \frac{19n - 1}{17n \sqrt{n}}$

converges or diverges.

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Step 1: Simplify the given series

For large $n$, the term $19n - 1$ behaves similarly to $19n$. So, the given term is asymptotically similar to:

$\frac{19n}{17n \sqrt{n}} = \frac{19}{17\sqrt{n}} = \frac{C}{\sqrt{n}}, \quad \text{where } C = \frac{19}{17}.$

Thus, we expect that the given series will behave like the p-series $\sum \frac{1}{\sqrt{n}}$, which is $\sum \frac{1}{n^{1/2}}$. We now compare our given series with the p-series.

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Step 2: Apply the Limit Comparison Test

The limit comparison test states that for two positive sequences $a_n$ and $b_n$, if:

$L = \lim_{n \to \infty} \frac{a_n}{b_n},$

and $L$ is a finite, positive number, then both series $\sum a_n$ and $\sum b_n$ will either both converge or both diverge.

Let:

$a_n = \frac{19n - 1}{17n \sqrt{n}}, \quad b_n = \frac{1}{\sqrt{n}}.$

Now, compute the limit:

$L = \lim_{n \to \infty} \frac{\frac{19n - 1}{17n \sqrt{n}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{19n - 1}{17n} = \lim_{n \to \infty} \frac{19 - \frac{1}{n}}{17}.$

As $n \to \infty$, $\frac{1}{n} \to 0$, so:

$L = \frac{19}{17}.$

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Step 3: Conclusion

Since $L = \frac{19}{17}$ is a finite, positive number, both the given series and the p-series $\sum \frac{1}{\sqrt{n}}$ will behave the same way.

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Step 4: Analyze the p-series

The series $\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}$ diverges because $p = 1/2 < 1$.

Thus, the given series:

$\sum_{n=2}^{\infty} \frac{19n - 1}{17n \sqrt{n}}$

also diverges by the limit comparison test.

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Do you want further details or have any questions?

Here are 5 related questions you might find useful:

1. How does the p-series test work, and why does it depend on $p$?
2. Can we use the ratio test instead of limit comparison in this case?
3. What is the intuition behind the limit comparison test?
4. What happens if the limit $L$ equals 0 or infinity in the comparison test?
5. How do we decide what series to compare with for limit comparison?

Tip: When dealing with series containing polynomial terms, always try to simplify the dominant terms for large $n$ to find a comparable series.