Problem 13 - Thomas Calculus

Question

Determine if the alternating series converges or diverges. Some of the series do not satisfy the conditions of the Alternating Series Test. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n}+1}{n+1} $$

Step-by-Step Solution

Answer

The series converges based on the Alternating Series Test.

Step 1: Identify the Alternating Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sqrt{n}+1}{n+1} \). It is alternating because of the term \((-1)^{n+1}\), which changes sign every term.

Step 2: Define the Positive Term Sequence

In this exercise, identify the sequence \(b_n = \frac{\sqrt{n}+1}{n+1}\). For the alternating series test, we need to verify if this sequence is decreasing and if the limit of \(b_n\) as \(n\) approaches infinity is zero.

Step 3: Check for Decreasing Sequence

To show \(b_n\) is decreasing, we should check if \(b_{n+1} \leq b_n\) for all \(n\). Let's compare \(b_n\) and \(b_{n+1}\). We have:\[ b_{n} = \frac{\sqrt{n}+1}{n+1} \text{ and } b_{n+1} = \frac{\sqrt{n+1}+1}{n+2} \]Show that \(b_{n} > b_{n+1}\) by cross-multiplying and simplifying the inequalities.

Step 4: Calculate the Limit of \(b_n\)

Calculate \( \lim_{n \to \infty} \frac{\sqrt{n}+1}{n+1} \). Divide the numerator and the denominator by \(n\):\[\lim_{n \to \infty} \frac{\sqrt{n}+1}{n+1} = \lim_{n \to \infty} \frac{\sqrt{n}/n + 1/n}{1 + 1/n} = \lim_{n \to \infty} \frac{1/\sqrt{n} + 1/n}{1 + 1/n} = 0\]Since both terms tend to zero as \(n\) approaches infinity.

Step 5: Apply the Alternating Series Test

The Alternating Series Test states that a series \(\sum (-1)^n a_n\) converges if \(a_n\) is positive, decreasing, and \(\lim_{n \to \infty} a_n = 0\). We have shown both conditions are met for \(b_n\), so the series converges.

Key Concepts

Series Convergence

When it comes to series convergence, it’s all about determining whether adding up an infinite number of terms results in a finite sum. In the context of alternating series, like the one we’re dealing with, the series converges if it passes the Alternating Series Test.

The alternating series in question is of the form \( \sum_{n=1}^{\infty} (-1)^{n+1} b_n \), where \( b_n \) is a positive sequence. To conclude that the series converges:

• The terms of \( b_n \) must be positive.
• \( b_n \) must eventually decline, becoming a decreasing sequence.
• The limit of \( b_n \) as \( n \) heads towards infinity must equal zero.

If these conditions hold, then the series converges. It’s like fitting pieces into a puzzle – if the pieces (or conditions) fit, then the big picture indicates convergence.

Limit Evaluation

Limit evaluation is crucial in judging whether a series qualifies for convergence. It's the process where we analyze how the sequence behaves as \( n \) becomes very large. For this exercise, we are looking at \( b_n = \frac{\sqrt{n} + 1}{n + 1} \), and our task is to find \( \lim_{n \to \infty} b_n \). The steps are straightforward:

• Simplify \( b_n \) by dividing both numerator and denominator by \( n \).
• This gives us \( \frac{\sqrt{n}/n + 1/n}{1 + 1/n} \), which simplifies further.

Now, as \( n \to \infty \), the terms \( 1/\sqrt{n} \) and \( 1/n \) in the numerator, and \( 1/n \) in the denominator, all approach zero. This evaluation tells us that \( \lim_{n \to \infty} b_n = 0 \).

The result affirms the second important criterion of the Alternating Series Test, ensuring that the overall infinite series converges.

Decreasing Sequences

A key characteristic required in the Alternating Series Test is that the sequence \( b_n \) should be decreasing, meaning each term is smaller than the one before it. To see if our sequence is decreasing:

• Define \( b_n = \frac{\sqrt{n} + 1}{n + 1} \) and \( b_{n+1} = \frac{\sqrt{n+1} + 1}{n+2} \).
• We want to establish whether \( b_{n+1} \leq b_n \).

This involves understanding the nature of the functions: As \( n \) grows larger, the numerator \( \sqrt{n} + 1 \) increases slowly due to the square root, while \( n + 1 \) in the denominator grows at a linear pace. Hence, \( b_n \) becomes smaller with increasing \( n \).

This pattern of decline guarantees the sequence is indeed decreasing, supporting the conditions for convergence as necessitated by the Alternating Series Test.