Problem 8 - Thomas Calculus

Question

In Exercises \(1-8,\) use the Direct Comparison Test to determine if each series converges or diverges. $$\sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{\sqrt{n^{2}+3}}$$

Step-by-Step Solution

Answer

The series diverges by the Direct Comparison Test.

Step 1: Understand the Series

The given series is \( \sum_{n=1}^{\infty} \frac{\sqrt{n}+1}{\sqrt{n^{2}+3}} \). We need to decide if this series converges or diverges using the Direct Comparison Test.

Step 2: Simplify the General Term

Simplify \( \frac{\sqrt{n}+1}{\sqrt{n^{2}+3}} \). For large \( n \), \( \sqrt{n^2+3} \approx n \), so the term simplifies approximately to \( \frac{\sqrt{n}+1}{n} \). As \( n \to \infty \), this becomes approximately \( \frac{1}{\sqrt{n}} \).

Step 3: Identify Comparison Series

Notice that \( \frac{1}{\sqrt{n}} \) is a well-known divergent p-series with \( p = \frac{1}{2} < 1 \). We will use \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) as our comparison series, which is known to diverge.

Step 4: Apply the Direct Comparison Test

For \( n \geq 1 \), observe:\[ \frac{\sqrt{n}+1}{\sqrt{n^{2}+3}} > \frac{1}{\sqrt{n}}. \] Since each term of the original series is greater than the corresponding term of the divergent comparison series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \), the original series also diverges by the Direct Comparison Test.

Key Concepts

series convergence

When we talk about series convergence, we are discussing whether the sum of an infinite series approaches a finite number. Imagine adding up all the terms of a series one by one. If this total eventually gets closer and closer to a certain number, the series is said to converge. Conversely, if the sum grows indefinitely, the series diverges.

Convergence is an essential concept because it helps us determine if a particular math recipe (series) provides a workable number at the end. To judge convergence, various tests are applied, such as the Direct Comparison Test, which we'll explore in detail. These tests compare terms in a series to those in known benchmark series to conclude about their convergence or divergence. In our example, a new series representation is studied by using common terms and comparing their properties.

p-series

A p-series is a specific kind of series characterized by its general term being in the form of \( \frac{1}{n^p} \), where \( n \) is a positive integer and \( p \) is a fixed positive constant. The convergence or divergence of a p-series depends critically on the exponent \( p \).

• If \( p > 1 \), the p-series converges. This means the sum of its infinite terms approaches a certain finite number.
• If \( p \leq 1 \), the p-series diverges. This implies the series adds up to infinity, never staying close to a fixed number.

In the exercise, the comparison series \( \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} \) is referred to as a p-series with \( p = 1/2 \). Since \( p = 1/2 < 1 \), this series diverges. Recognizing the p-series type helps to quickly determine the behavior of infinite sequences.

divergence test

The Divergence Test is a simple yet powerful tool for analyzing series. It tells us that if the limit of a series' term does not equal zero as the term number approaches infinity, then the series definitely diverges. In other words, if the terms aren't getting closer to zero, the entire series has no chance of converging to a finite sum.

This test is especially handy as a preliminary check when working with unknown series. However, failing the divergence test doesn’t guarantee convergence; it merely rules out the possibility of divergence. In our example series, the individual test was not needed, as the Direct Comparison Test demonstrated divergence by showing the terms exceed those of a known divergent p-series. That comparison allowed us to conclude that the series diverges without directly using the divergence test.