Problem 13 - Thomas Calculus

Question

In Exercises \(1 - 14 ,\) 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 by the Alternating Series Test.

Step 1: Identify the General Term

The given alternating series is \( \sum _{n=1} ^{\infty} (-1)^{n+1} \frac{\sqrt{n} + 1}{n + 1} \). The general term \( a_n \) can be expressed as \( a_n = \frac{\sqrt{n} + 1}{n + 1} \).

Step 2: Check Non-increasing Condition

For the alternating series test, the terms \( a_n \) need to be non-increasing. Evaluate \( a_{n+1} = \frac{\sqrt{n+1} + 1}{n+2} \) and compare it with \( a_n \). If \( a_{n+1} \leq a_n \) for sufficiently large \( n \), then this condition is satisfied.Comparing \( a_{n+1} \) and \( a_n \):\[\frac{\sqrt{n+1} + 1}{n+2} \leq \frac{\sqrt{n} + 1}{n+1}\]Simplifying is non-trivial without calculus, but qualitatively, since the denominator grows faster relative to the numerator, \( a_n \) is decreasing for large \( n \).

Step 3: Evaluate the Limit of the Terms

Next, determine if \( \lim_{{n \to \infty}} a_n = 0 \). Calculate: \[\lim_{n \to \infty} \frac{\sqrt{n} + 1}{n + 1} = \lim_{n \to \infty} \left( \frac{\sqrt{n}}{n} + \frac{1}{n+1} \right) \] \[= \lim_{n \to \infty} \left( \frac{1}{\sqrt{n}} + \frac{1}{n+1} \right) = 0 + 0 = 0\]Thus, \( \lim_{n \to \infty} a_n = 0 \) is satisfied.

Step 4: Apply the Alternating Series Test

We have verified that the terms \( a_n \) are non-increasing and \( \lim_{n \to \infty} a_n = 0 \). By the Alternating Series Test, the series converges.

Key Concepts

Convergence and Divergence

When dealing with an alternating series, understanding whether it converges or diverges is crucial. **Convergence** of a series means that as you sum more and more terms, the total reaches a limit, a fixed number where the sum "settles." If a series converges, it means adding more terms changes the sum less and less, eventually having little to no effect.
On the other hand, a **divergent** series has sums that grow without bound or do not approach any limit.

• For alternating series like the one in the exercise, where terms change their sign between positive and negative, specific tests can be applied to determine convergence.
• The **Alternating Series Test** checks two main conditions: whether the terms' absolute values are decreasing, and if the terms tend to zero as they progress.

If these conditions are met, the series converges. This is exactly what was demonstrated in the exercise, leading to the conclusion that the series converges.

General Term of a Series

The general term of a series is immensely important as it represents each term in the series. Understanding it allows us to explore the series' properties. In the exercise, the given alternating series can be written as:
\[(-1)^{n+1} rac{\sqrt{n} + 1}{n + 1}\]

• The general term here is:
  \(a_n = \frac{\sqrt{n} + 1}{n + 1}\)
• The alternating factor \((-1)^{n+1}\) makes each subsequent term positive if the prior was negative, and vice versa.

To apply the Alternating Series Test, focus on \(a_n\) ignoring the negative sign flipping element.
Understanding the behavior of \(a_n\) helps establish whether the series decreases and approaches zero, key factors for determining the series' convergence.

Limit of a Sequence

The limit of a sequence is a foundational idea in calculus and series analysis. It tells us what value a sequence approaches as it progresses into infinity. In our exercise, finding \(\lim_{n \to \infty} a_n\) is a crucial step. Let's consider why.

• For convergence in alternating series, not only must terms decrease in absolute value, but they also need to approach zero.
• Calculating:\[\lim_{n \to \infty} \left( \frac{\sqrt{n} + 1}{n + 1} \right) = \lim_{n \to \infty} \left( \frac{1}{\sqrt{n}} + \frac{1}{n+1} \right) = 0\] shows that this condition is satisfied.

Reaching a limit of zero effectively suggests that as you go further along the sequence, each term adds less to the series, supporting its convergence.