Comparison Test: Convergence Of ∑ (n^(19/9) / (n^3 - N))

Hey guys! Today, we're diving into the fascinating world of infinite series and tackling a convergence problem using the Comparison Test. Specifically, we're going to determine whether the infinite series ∑[n=2 to ∞] (n^(19/9) / (n^3 - n)) converges or diverges. So, buckle up and let's get started!

Understanding the Comparison Test

Before we jump into the problem, let's quickly recap what the Comparison Test is all about. The Comparison Test is a powerful tool for determining the convergence or divergence of an infinite series by comparing it to another series whose convergence behavior is already known. There are essentially two versions of the test:

1. Direct Comparison Test:
   • If 0 ≤ aₙ ≤ bₙ for all n, and ∑ bₙ converges, then ∑ aₙ also converges.
   • If aₙ ≥ bₙ ≥ 0 for all n, and ∑ bₙ diverges, then ∑ aₙ also diverges.
2. Limit Comparison Test:
   • If aₙ > 0 and bₙ > 0 for all n, and lim (n→∞) (aₙ / bₙ) = c, where 0 < c < ∞, then either both series ∑ aₙ and ∑ bₙ converge, or both diverge.

In essence, the Direct Comparison Test lets us sandwich our series between two others, while the Limit Comparison Test allows us to compare the long-term behavior of two series. For this problem, we'll be leveraging the Limit Comparison Test, as it's often more flexible when dealing with algebraic expressions.

Applying the Comparison Test to Our Series

Now, let's get back to our series: ∑[n=2 to ∞] (n^(19/9) / (n^3 - n)). The key to using the Comparison Test lies in choosing the right series to compare with. We need a series whose convergence behavior we already know and that behaves similarly to our given series for large values of n. Let's analyze the dominant terms in our series.

For large n, the term n in the denominator becomes insignificant compared to n³. Therefore, n³ - n behaves approximately like n³. This means our series behaves roughly like n^(19/9) / n³ = n^(19/9 - 3) = n^(19/9 - 27/9) = n^(-8/9) = 1/n^(8/9). This suggests we should compare our series with the p-series ∑ 1/n^(8/9).

Choosing a Comparison Series

Our comparison series will be ∑[n=2 to ∞] 1/n^(8/9). This is a p-series with p = 8/9. Recall that a p-series ∑ 1/n^p converges if p > 1 and diverges if p ≤ 1. Since 8/9 < 1, our comparison series ∑ 1/n^(8/9) diverges.

Performing the Limit Comparison Test

Now, let's apply the Limit Comparison Test. We need to compute the limit:

lim (n→∞) [(n^(19/9) / (n³ - n)) / (1/n^(8/9))]

This looks a bit intimidating, but we can simplify it. Dividing by a fraction is the same as multiplying by its reciprocal, so we get:

lim (n→∞) [(n^(19/9) / (n³ - n)) * n^(8/9)]

lim (n→∞) [n^(19/9 + 8/9) / (n³ - n)]

lim (n→∞) [n^(27/9) / (n³ - n)]

lim (n→∞) [n³ / (n³ - n)]

Now, we can divide both the numerator and the denominator by n³:

lim (n→∞) [1 / (1 - 1/n²)]

As n approaches infinity, 1/n² approaches 0, so the limit becomes:

lim (n→∞) [1 / (1 - 0)] = 1

Interpreting the Result

We found that the limit is 1, which is a finite number greater than 0 (0 < 1 < ∞). This satisfies the condition for the Limit Comparison Test. Since our comparison series ∑ 1/n^(8/9) diverges (because it's a p-series with p = 8/9 ≤ 1), the Limit Comparison Test tells us that our original series, ∑[n=2 to ∞] (n^(19/9) / (n³ - n)), also diverges.

Conclusion

Therefore, using the Limit Comparison Test, we've determined that the infinite series ∑[n=2 to ∞] (n^(19/9) / (n³ - n)) diverges. The key was to identify a suitable comparison series – in this case, a p-series – and then carefully evaluate the limit of the ratio of the terms. Remember, guys, practice makes perfect, so keep tackling those series problems!

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Further Exploration

If you're interested in learning more about convergence tests, here are a few topics you might want to explore:

• Ratio Test: Another powerful test for determining convergence, especially useful for series involving factorials.
• Root Test: Similar to the Ratio Test, but often easier to apply in certain situations.
• Alternating Series Test: Specifically designed for series with alternating signs.
• Integral Test: Connects the convergence of a series to the convergence of an integral.

Understanding these different tests will give you a more complete toolkit for analyzing infinite series. Keep practicing, guys, and you'll become series experts in no time!

Practice Problems

Want to put your newfound knowledge to the test? Try these practice problems:

1. Use the Comparison Test to determine whether the series ∑[n=1 to ∞] (1 / (n² + n)) converges or diverges.
2. Use the Limit Comparison Test to determine whether the series ∑[n=1 to ∞] (√(n) / (n² + 1)) converges or diverges.
3. Determine the convergence or divergence of the series ∑[n=1 to ∞] (n / (n³ + 1)) using an appropriate test.

These problems will help you solidify your understanding of the Comparison Test and other convergence techniques. Good luck, guys!