Problem 10 - Thomas Calculus

Question

In Exercises \(9-16,\) use the Limit Comparison Test to determine if each series converges or diverges. $$ \begin{array}{l}{\sum_{n=1}^{\infty} \sqrt{\frac{n+1}{n^{2}+2}}} \\\ {\left(\text {Hint} : \text { Limit Comparison with } \sum_{n=1}^{\infty}(1 / \sqrt{n})\right)}\end{array} $$

Step-by-Step Solution

Answer

The series diverges by the Limit Comparison Test with \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \).

Step 1: Identify the Given Series

We are given the series \( \sum_{n=1}^{\infty} \sqrt{\frac{n+1}{n^2+2}} \). We need to determine whether this series converges or diverges using the Limit Comparison Test.

Step 2: Determine the Comparison Series

The hint suggests comparing with the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), which is a \( p \)-series with \( p = \frac{1}{2} \). Since \( p < 1 \), the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) diverges.

Step 3: Apply the Limit Comparison Test

The Limit Comparison Test states that for non-negative series \( a_n \) and \( b_n \), if \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) where \( 0 < c < \infty \), then both series either converge or diverge. Set \( a_n = \sqrt{\frac{n+1}{n^2+2}} \) and \( b_n = \frac{1}{\sqrt{n}} \).

Step 4: Find the Limit

Calculate the limit: \[ \lim_{n \to \infty} \frac{\sqrt{\frac{n+1}{n^2+2}}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \sqrt{\frac{n(n+1)}{n^2+2}}. \] Simplifying, this becomes: \[ \lim_{n \to \infty} \sqrt{\frac{n^2+n}{n^2+2}} = \sqrt{\lim_{n \to \infty} \frac{1 + \frac{1}{n}}{1 + \frac{2}{n^2}}}. \] As \( n \to \infty \), the limit inside the square root simplifies to \( \frac{1}{1} = 1 \). Therefore, the limit is \( \sqrt{1} = 1 \).

Step 5: Conclude with Limit Comparison Test

Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 1 \) and \( 0 < 1 < \infty \), by the Limit Comparison Test, the given series \( \sum_{n=1}^{\infty} \sqrt{\frac{n+1}{n^2+2}} \) diverges because the comparison series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a valuable tool when evaluating the convergence or divergence of a series. It works best when you have a series that doesn't immediately fit the criteria for basic tests. This test allows us to compare the given series to another one that is easier to analyze.
To use the Limit Comparison Test, follow these steps:

• Identify the series you need to evaluate, known as the target series.
• Choose a series to compare with, ideally one whose convergence properties are already known. This is often called the comparison series.
• Calculate the limit of the ratio of the two series' terms as they approach infinity: \[ \lim_{n \to \infty} \frac{a_n}{b_n} \] where \(a_n\) are terms from the target series, and \(b_n\) are terms from the comparison series.
• If the limit \(c\) is positive and finite, i.e., \(0 < c < \infty\), then both the target and the comparison series either both converge or both diverge.

In our example, we compared \(a_n = \sqrt{\frac{n+1}{n^2+2}}\) with \(b_n = \frac{1}{\sqrt{n}}\), a known challenging series.

Divergent Series

A divergent series is one that does not converge; its terms do not approach a fixed limit as they extend towards infinity. Understanding divergent series is essential in mathematical analysis, as it often relates to the behavior of functions and their integrals.
When dealing with series, particularly in calculus, we look for signs of convergence or divergence. If a series diverges, it means its sum grows without bound or oscillates indefinitely. In the context of the given exercise, the series \( \sum_{n=1}^{\infty} \sqrt{\frac{n+1}{n^2+2}} \) was evaluated using the Limit Comparison Test. We observed it against the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), which is known to diverge because of the nature of its terms resembling a harmonic-like series with denominator terms not large enough to enforce convergence. Thus, by the test, our target series also diverges.

P-Series Test

The P-Series Test is a fundamental concept when discussing series in mathematics. A p-series is of the form:
\[\sum_{n=1}^{\infty} \frac{1}{n^p}\]Where \(p\) is a real number. The convergence of such a series depends entirely on the value of \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

In our original exercise, we examined the series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), equating it to a p-series where \(p = \frac{1}{2}\). Since \(p < 1\), it is a divergent p-series. This quickly helps determine how the target series, given in the exercise, behaves when compared using the Limit Comparison Test. Thus, knowledge about p-series significantly aids in evaluating more complex series.