Convergence And Divergence Analysis Of Infinite Series

In the realm of mathematical analysis, determining whether an infinite series converges or diverges is a fundamental problem. This article delves into the convergence and divergence analysis of two specific series. We will employ various convergence tests, such as the Limit Comparison Test, to rigorously assess the behavior of each series. Understanding the convergence or divergence of series is crucial in many areas of mathematics, including calculus, differential equations, and complex analysis.

(a) Analyzing the Convergence of {\sum rac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}}

Applying the Limit Comparison Test

To determine the convergence or divergence of the series {\sum rac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}}, we can utilize the Limit Comparison Test. This test involves comparing the given series with a simpler series whose convergence behavior is already known. A suitable comparison series in this case is {\sum rac{\sqrt{n^2}}{\sqrt{n^5}} = \sum rac{n}{n^{5/2}} = \sum rac{1}{n^{3/2}}}. This comparison series is a p-series with {p = rac{3}{2} > 1}, which is known to converge.

The Limit Comparison Test states that if the limit of the ratio of the terms of two series is a finite, non-zero number, then both series either converge or diverge together. Let's compute the limit:

{ \lim_{n \to \infty} rac{\frac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} rac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}} \cdot n^{3/2} = \lim_{n \to \infty} rac{n^{3/2} \sqrt{n^2 - 1}}{\sqrt{n^5 + 1}} }

To simplify this limit, we divide both the numerator and the denominator by appropriate powers of ${n}$:

{ \lim_{n \to \infty} rac{\sqrt{n^5 - n^3}}{\sqrt{n^5 + 1}} = \lim_{n \to \infty} \sqrt{\frac{n^5 - n^3}{n^5 + 1}} = \lim_{n \to \infty} \sqrt{\frac{1 - rac{1}{n^2}}{1 + rac{1}{n^5}}} }

As ${n \to \infty}$, the terms {rac{1}{n^2}} and {rac{1}{n^5}} approach 0, so the limit becomes:

${ \lim_{n \to \infty} \sqrt{\frac{1 - 0}{1 + 0}} = \sqrt{1} = 1 }$

Since the limit is 1, which is a finite, non-zero number, the given series {\sum rac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}} behaves the same as the comparison series {\sum rac{1}{n^{3/2}}\. Because the comparison series converges (as it is a p-series with \(p = rac{3}{2} > 1}), we conclude that the given series also converges.

Conclusion for Part (a)

In conclusion, by applying the Limit Comparison Test with the convergent p-series {\sum rac{1}{n^{3/2}}\, we have rigorously shown that the series \(\sum rac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}} converges. This determination highlights the effectiveness of the Limit Comparison Test in analyzing the convergence behavior of complex series by relating them to simpler, well-understood series.

(b) Analyzing the Convergence of {\sum rac{1}{n} an{\frac{\pi}{n}}}

Applying the Limit Comparison Test

For the series {\sum rac{1}{n} an{\frac{\pi}{n}}}, we again employ the Limit Comparison Test to determine its convergence or divergence. To effectively apply this test, we need to select a suitable comparison series. Given the presence of the tangent function, it is advantageous to consider the small-angle approximation of the tangent function. Specifically, for small angles ${x}$, ${\tan{x} \approx x}$. Thus, we can approximate ${\tan{\frac{\pi}{n}} \approx \frac{\pi}{n}}$ for large ${n}$.

Using this approximation, we can compare the given series with the series {\sum rac{1}{n} rac{\pi}{n} = \sum rac{\pi}{n^2}}. Since ${\pi}$ is a constant, this series is essentially a constant multiple of the series {\sum rac{1}{n^2}}, which is a p-series with ${p = 2}$. This p-series is known to converge because ${p > 1}$.

Now, we apply the Limit Comparison Test by computing the limit of the ratio of the terms of the two series:

{ \lim_{n \to \infty} rac{\frac{1}{n} an{\frac{\pi}{n}}}{\frac{\pi}{n^2}} = \lim_{n \to \infty} rac{1}{n} an{\frac{\pi}{n}} \cdot rac{n^2}{\pi} = \lim_{n \to \infty} rac{n}{\pi} an{\frac{\pi}{n}} }

To evaluate this limit, we can use the substitution {x = rac{\pi}{n}}. As ${n \to \infty}$, ${x \to 0}$. The limit then becomes:

{ \lim_{x \to 0} rac{\pi}{x} \frac{\tan{x}}{\pi} = \lim_{x \to 0} rac{\tan{x}}{x} }

This is a well-known limit in calculus. We can evaluate it using L'Hôpital's Rule or by recognizing the standard limit. Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to ${x}$:

{ \lim_{x \to 0} rac{\sec^2{x}}{1} = \sec^2{0} = 1 }

Since the limit is 1, which is a finite, non-zero number, the given series {\sum rac{1}{n} an{\frac{\pi}{n}}} behaves the same as the comparison series {\sum rac{\pi}{n^2}}. Because the comparison series converges (as it is a constant multiple of the convergent p-series {\sum rac{1}{n^2}}), we conclude that the given series also converges.

Conclusion for Part (b)

In conclusion, through the application of the Limit Comparison Test, and by leveraging the small-angle approximation of the tangent function, we have conclusively demonstrated that the series {\sum rac{1}{n} an{\frac{\pi}{n}}} converges. This analysis underscores the importance of selecting appropriate comparison series and utilizing trigonometric approximations in the convergence testing of series.

In summary, we have meticulously analyzed the convergence behavior of two distinct series. For the series {\sum rac{\sqrt{n^2 - 1}}{\sqrt{n^5 + 1}}}, we employed the Limit Comparison Test with the convergent p-series {\sum rac{1}{n^{3/2}}} to establish its convergence. Similarly, for the series {\sum rac{1}{n} an{\frac{\pi}{n}}}, we again used the Limit Comparison Test, comparing it with the convergent series {\sum rac{\pi}{n^2}}, which is a multiple of the p-series {\sum rac{1}{n^2}}. Through these analyses, we have shown that both series converge, reinforcing the significance and versatility of the Limit Comparison Test in the determination of convergence or divergence of infinite series. The understanding of these tests is vital for students and professionals in mathematics and related fields.