Convergence Test: Series Σ 1/(n³-216) Analysis

Let's dive into determining whether the series $\sum_{n=7}^{\infty} \frac{1}{n^3-216}$ converges or diverges. This involves employing one of several convergence tests to analyze the behavior of the series as $n$ approaches infinity. We will explore the characteristics of the given series and strategically apply an appropriate convergence test to reach a definitive conclusion.

Understanding the Series

The given series is $\sum_{n=7}^{\infty} \frac{1}{n^3-216}$. A quick observation reveals that the denominator is $n^3 - 216$, which can be factored as $n^3 - 6^3$. Factoring this difference of cubes gives $(n-6)(n^2 + 6n + 36)$. Notice that the series starts at $n=7$, so we avoid any division by zero. The terms of the series are positive and decrease as $n$ increases. The structure of the series suggests that a comparison test could be effective.

Choosing the Right Convergence Test

Several tests are available to determine convergence or divergence, including:

• n-th Term Test
• Geometric Series Test
• p-Series Test
• Telescoping Series Test
• Integral Test
• Comparison Tests (Direct Comparison Test and Limit Comparison Test)

Given the form of our series, the Direct Comparison Test or the Limit Comparison Test appear to be the most promising. Let's explore both to see which one is more suitable.

Direct Comparison Test

For the Direct Comparison Test, we need to find a simpler series to compare with our given series. We want to compare $\frac{1}{n^3 - 216}$ with a series of the form $\frac{1}{n^p}$. As $n$ becomes large, the $-216$ becomes less significant, so we can compare it with $\frac{1}{n^3}$.

However, we need to ensure that $\frac{1}{n^3 - 216} > \frac{1}{n^3}$ for all $n \geq 7$. This inequality holds because $n^3 > n^3 - 216$, and taking the reciprocal reverses the inequality. The series $\sum_{n=7}^{\infty} \frac{1}{n^3}$ is a $p$-series with $p = 3 > 1$, which converges. Unfortunately, since our series is greater than a convergent series, the Direct Comparison Test is inconclusive in this case.

Limit Comparison Test

The Limit Comparison Test involves taking the limit of the ratio of our series term and the term of a known convergent or divergent series. Again, let's compare $\frac{1}{n^3 - 216}$ with $\frac{1}{n^3}$. We calculate the limit:

$$\lim_{n \to \infty} \frac{\frac{1}{n^3 - 216}}{\frac{1}{n^3}} = \lim_{n \to \infty} \frac{n^3}{n^3 - 216} = \lim_{n \to \infty} \frac{1}{1 - \frac{216}{n^3}} = 1$$

Since the limit is a finite number greater than 0 (specifically, 1), the Limit Comparison Test tells us that $\sum_{n=7}^{\infty} \frac{1}{n^3 - 216}$ behaves the same way as $\sum_{n=7}^{\infty} \frac{1}{n^3}$.

The p-Series Test

The series $\sum_{n=7}^{\infty} \frac{1}{n^3}$ is a $p$-series with $p = 3$. According to the p-Series Test, a series of the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$. In our case, $p = 3 > 1$, so the series $\sum_{n=7}^{\infty} \frac{1}{n^3}$ converges.

Conclusion

Since $\sum_{n=7}^{\infty} \frac{1}{n^3}$ converges and the limit comparison test showed that $\sum_{n=7}^{\infty} \frac{1}{n^3-216}$ behaves the same way, we conclude that the series $\sum_{n=7}^{\infty} \frac{1}{n^3-216}$ also converges. We used the Limit Comparison Test in conjunction with the p-Series Test to reach this conclusion.

Additional Insights and Considerations

While the Limit Comparison Test provided a straightforward solution, it's valuable to consider why other tests might not have been as effective or direct. The n-th Term Test, for instance, only tells us if a series diverges when the limit of the terms does not approach zero. In this case, the terms clearly approach zero, so the test is inconclusive.

The Geometric Series Test is not applicable because the series is not in the form of a geometric series. The Telescoping Series Test would require the series to be expressed as a difference of consecutive terms, which is not apparent here. The Integral Test could be used, but it often involves more complicated integration than the Limit Comparison Test required.

Therefore, the Limit Comparison Test was the most efficient and direct method for determining the convergence of this series. Understanding the strengths and weaknesses of each test allows for strategic problem-solving in convergence and divergence analysis.

In summary, when dealing with series, it's essential to analyze the structure of the series and choose the convergence test that best fits its form. The Limit Comparison Test is particularly useful when the series can be compared to a known p-series or another series with a known convergence behavior.

Further Exploration

For a deeper understanding of convergence tests and series analysis, explore resources like Khan Academy's Calculus section on series. This resource provides comprehensive explanations, examples, and practice problems to enhance your knowledge of series convergence and divergence.