Problem 18 - Calculus: Early Transcendentals

Question

Determine whether the series converges or diverges. \( \displaystyle \sum_{n = 1}^{\infty} \frac {2}{\sqrt n + 2} \)

Step-by-Step Solution

Answer

The series diverges.

Step 1: Set up the Series

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

Step 2: Propose a Test for Convergence

For this series, considering its form, we can use the Limit Comparison Test. We'll compare it with the series \( \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} \), which is a \( p \)-series with \( p = \frac{1}{2} \). Recall that a \( p \)-series converges only if \( p > 1 \). Therefore, \( \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} \) diverges.

Step 3: Apply the Limit Comparison Test

Compute the limit: \[ \lim_{n \to \infty} \frac{\frac{2}{\sqrt{n} + 2}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{2}{\sqrt{n} + 2} \cdot \sqrt{n} = \lim_{n \to \infty} \frac{2\sqrt{n}}{\sqrt{n} + 2}. \]

Step 4: Simplify and Evaluate the Limit

Simplify the expression: \[ \lim_{n \to \infty} \frac{2\sqrt{n}}{\sqrt{n} + 2} = \lim_{n \to \infty} \frac{2\sqrt{n}}{\sqrt{n}(1 + \frac{2}{\sqrt{n}})} = \lim_{n \to \infty} \frac{2}{1 + \frac{2}{\sqrt{n}}}. \] As \( n \to \infty \), \( \frac{2}{\sqrt{n}} \to 0 \), so the limit becomes \( \frac{2}{1} = 2 \).

Step 5: Conclude Based on the Limit Comparison Test

Since \( \lim_{n \to \infty} \frac{a_n}{b_n} = 2 \) and \( 2 > 0 \), by the Limit Comparison Test, both series \( \sum_{n = 1}^{\infty} \frac{2}{\sqrt{n} + 2} \) and \( \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} \) diverge or converge together. Since \( \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} \) diverges, we can conclude that \( \sum_{n = 1}^{\infty} \frac{2}{\sqrt{n} + 2} \) also diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool used in calculus to determine the convergence or divergence of a series. It works on the principle of comparing two series whose terms \(a_n\) and \(b_n\) are positive and similar in nature.

To apply the Limit Comparison Test, you follow these steps:

• Identify two series: one being the target series \(\sum a_n\) whose convergence you wish to determine, and another known series \(\sum b_n\) with similar terms.
• Calculate the limit \L = \lim_{n \to \infty} \frac{a_n}{b_n}\.
• If L is a positive finite number (L > 0), then both series either converge or diverge together.

In the original exercise, we applied this test by comparing the given series \(\sum_{n=1}^{\infty} \frac{2}{\sqrt{n} + 2}\) with the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\). By calculating the limit, we found it equaled 2, indicating both series behave in the same way.

p-series

A p-series is a fundamental type of infinite series, often written in the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). The value of \(p\) determines whether the series converges or diverges.

Here's the rule of thumb for p-series:

• If \(p > 1\), the series converges.
• If \(0 < p \leq 1\), the series diverges.

This is crucial for analyzing convergence in relation to other series, as shown in the original solution. The comparison series in the exercise, \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} = \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}\), is a classic example of a diverging p-series because \(\frac{1}{2} \leq 1\). This characteristic was vital to concluding the nature of the original series.

divergence

In calculus, a series is said to diverge when its sequence of partial sums doesn't approach a finite limit as n approaches infinity.

There are multiple ways a series can diverge:

• The series grows indefinitely, without bound.
• The series oscillates or doesn't settle to a specific value.

In this context, divergence implies that the sum is not finite and the series does not stabilize. For example, in our exercise, the series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) demonstrates divergence by growing indefinitely.

When using the Limit Comparison Test, discovering that a comparison series diverges, like in this case, allows us to conclude similarly about the target series. The series \(\sum_{n=1}^{\infty} \frac{2}{\sqrt{n} + 2}\) diverges together with its companion series.