Problem 33 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation} \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} \end{equation}

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with \( \sum \frac{1}{n^2} \).

Step 1: Identify the Type of Series

The series given is \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} \). This is an infinite series whose general term is \( a_n = \frac{1}{n \sqrt{n^{2}-1}} \). We need to determine whether this series converges or diverges.

Step 2: Compare with a Known Series

We notice that \( \frac{1}{n \sqrt{n^{2}-1}} \) is similar to \( \frac{1}{n^2} \) for large \( n \). We can use the Limit Comparison Test with the series \( \sum_{n=2}^{\infty} \frac{1}{n^2} \), a p-series with \( p = 2 > 1 \), which is known to converge.

Step 3: Apply Limit Comparison Test

Define \( b_n = \frac{1}{n^2} \). Compute the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \). We have:\[L = \lim_{n \to \infty} \frac{\frac{1}{n \sqrt{n^{2}-1}}}{\frac{1}{n^2}} = \lim_{n \to \infty} \frac{n^2}{n \sqrt{n^{2} - 1}} = \lim_{n \to \infty} \frac{n}{\sqrt{n^{2} - 1}}\]Simplify the expression:\[L = \lim_{n \to \infty} \frac{n}{\sqrt{n^{2}(1 - \frac{1}{n^{2}})}} = \lim_{n \to \infty} \frac{n}{n \sqrt{1 - \frac{1}{n^{2}}}} = \lim_{n \to \infty} \frac{1}{\sqrt{1 - \frac{1}{n^{2}}}}\]As \( n \to \infty \), \( 1 - \frac{1}{n^2} \to 1 \), so \( \sqrt{1 - \frac{1}{n^{2}}} \to 1 \), hence \( L = 1 \).

Step 4: Conclude Convergence

Since \( L = 1 \) and \( 0 < L < \infty \), by the Limit Comparison Test, the series \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2}-1}} \) converges, since \( \sum_{n=2}^{\infty} \frac{1}{n^2} \) converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a useful tool when dealing with infinite series, especially when the series in question resembles another series whose behavior is already known. It essentially helps us to determine the convergence or divergence of a given series by comparing it to a second series that we already understand or know well.

So how does it work? Let's get into it! Imagine you have a series with terms denoted as \( a_n \) and you choose another series with terms \( b_n \) that is similar in nature to \( a_n \). The steps to apply the Limit Comparison Test are as follows:

• First, compute the limit \( L = \lim_{{n \to \infty}} \frac{{a_n}}{{b_n}} \).
• If \( L \) is a positive finite number, meaning \( 0 < L < \infty \), both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.
• If \( L \) equals zero or infinity, the test is inconclusive; thus, you may want to try a different analysis method.

In the original exercise, we compared the given series to a well-known convergent series \( \sum \frac{1}{n^2} \), finding that the limit \( L \) was 1, which suggests that both series converge.

Using the Limit Comparison Test requires a keen eye for spotting series that can be effectively compared, allowing you to save time and effort in determining convergence.

p-series

A p-series is a specific kind of infinite series that takes the form \( \sum \frac{1}{n^p} \), where \( p \) is a positive constant and \( n \) starts from some integer value onward.
Why are they important? Well, p-series are handy because their convergence behavior is well-known:

• When \( p > 1 \), the p-series \( \sum \frac{1}{n^p} \) converges.
• When \( p \leq 1 \), the p-series \( \sum \frac{1}{n^p} \) diverges.

P-series serve as a benchmark for comparing other series, which is why they are often used in tests like the Limit Comparison Test.

In the exercise solution, the series \( \sum \frac{1}{n^2} \), having \( p = 2 \), was used as a basis for comparison because it is a p-series with \( p > 1 \). Knowing the convergence of \( \sum \frac{1}{n^2} \) provided a solid foundation to apply the Limit Comparison Test effectively.

Infinite Series

An infinite series is simply the sum of an infinite sequence of terms. Mathematically, it is expressed as \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) are the terms of the series.

Infinite series are fascinating because they encapsulate endless summation which can either converge to a single finite value, or diverge without bound. Understanding whether an infinite series converges or diverges is crucial in many areas of mathematics, science, and engineering.

When working with infinite series, we often assess convergence using tests like:

• Divergence Test
• Ratio Test
• Root Test
• Limit Comparison Test

Choosing the right test necessitates a good understanding of the series structure. In the original exercise, we dealt with a series \( \sum_{n=2}^{\infty} \frac{1}{n \sqrt{n^{2} - 1}} \), which requires a careful check for convergence.

Infinite series feature prominently in integral calculus and differential equations, making them a cornerstone of advanced mathematics. Learning how to work with and understand them can greatly enhance problem-solving skills across disciplines.