Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{n^{2}+1} $$

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with \( \sum \frac{1}{n^{3/2}} \).

Step 1: Identify the Series

The series in question is \( \sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{n^{2}+1} \). We need to determine if this series converges or diverges.

Step 2: Choose a Convergence Test

Since the series involves square roots and polynomials, the Limit Comparison Test or the Direct Comparison Test are suitable options to analyze the series' behavior at infinity.

Step 3: Select a Comparison Series

The term can be simplified for large \( n \) by approximating \( \sqrt{n+1} \approx \sqrt{n} \). Compare it with a simpler series \( \sum \frac{\sqrt{n}}{n^2} = \sum \frac{1}{n^{3/2}} \). This comparison series is a p-series with \( p = 3/2 > 1 \), which is known to converge.

Step 4: Apply the Limit Comparison Test

To use the Limit Comparison Test, evaluate \( \lim_{n \to \infty} \frac{\frac{\sqrt{n+1}}{n^2+1}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{\sqrt{n+1} \cdot n^{3/2}}{n^2+1} \). Simplify the limit: \[ \lim_{n \to \infty} \frac{\sqrt{n+1} \cdot \sqrt{n^3}}{n^2+1} = \lim_{n \to \infty} \frac{n^{3/2}\sqrt{1+\frac{1}{n}}}{n^2(1+\frac{1}{n^2})} = \lim_{n \to \infty} \frac{n^{3/2}}{n^2} = \lim_{n \to \infty} \frac{1}{n^{1/2}} = 0 \]The limit is non-zero and finite.

Step 5: Conclusion of the Test

From the Limit Comparison Test, since \( \sum \frac{1}{n^{3/2}} \) converges and the limit is a finite positive number, \( \sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{n^{2}+1} \) also converges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a method used to determine if a series, \( \sum a_n \), converges or diverges by comparing it to another series, \( \sum b_n \), whose convergence properties are already known. This test is especially useful when dealing with series that are complex or difficult to analyze directly.
For the test to be applicable:

• Both \( a_n \) and \( b_n \), must be positive for all sufficiently large \( n \).
• Compute the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = L \).

If \( L \) is a positive finite number, then both series \( \sum a_n \) and \( \sum b_n \) either converge or diverge together.

In the given problem, we compared \( \frac{\sqrt{n+1}}{n^2+1} \) to \( \frac{1}{n^{3/2}} \). By evaluating the limit, it was shown to be finite and non-zero, suggesting the behavior of both series is similar at infinity.

p-series

The p-series is one of the standard types of series used in convergence testing and is defined as \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence of a p-series depends entirely on the value of \( p \):

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

For this exercise, the chosen comparison series \( \sum \frac{1}{n^{3/2}} \) is a p-series with \( p = 1.5 \). Since \( p = 1.5 > 1 \), this series converges. This fact is crucial in using the Limit Comparison Test, as the convergence of the known series \( \sum \frac{1}{n^{3/2}} \) helped conclude the behavior of the original series.

Convergent Series

A convergent series is a series whose terms approach a sum that is finite as more and more terms are added. In simpler terms, as you add up the terms of the series, they get closer and closer to a certain number. Convergent series are essential in calculus and analysis, as they rely on understanding series behavior to compute limits, integrals, and more.

In the context of this exercise, we concluded that the series \( \sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{n^2+1} \) is convergent by applying the Limit Comparison Test with a known convergent p-series \( \sum \frac{1}{n^{3/2}} \). Since the limit of their ratio was positive and finite, the convergence of the known series was transferred to the given series.

Recognizing whether a series converges is a crucial skill in mathematics, as it informs on the behavior of more complex functions and their integrals.