Problem 14 - 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. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{3 \sqrt{n+1}}{\sqrt{n}+1} $$

Step-by-Step Solution

Answer

The series diverges because the terms \( a_n \) do not tend to zero as \( n \to \infty \).

Step 1: Identifying the Series Components

The given series is \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{3 \sqrt{n+1}}{\sqrt{n}+1} \). The series alternates because of the \((-1)^{n+1}\) term.

Step 2: Checking the Alternating Series Test Conditions - Condition 1

An alternating series \( \sum (-1)^{n+1} a_n \) converges if the terms \( a_n \) are positive, decreasing, and tend to zero as \( n \to \infty \). We define \( a_n = \frac{3 \sqrt{n+1}}{\sqrt{n}+1} \), which is indeed positive for all \( n \geq 1 \).

Step 3: Checking the Alternating Series Test Conditions - Condition 2

Next, we examine whether \( a_n \) is decreasing. To check this, we compare \( a_n \) and \( a_{n+1} \):\[ a_{n+1} = \frac{3 \sqrt{n+2}}{\sqrt{n+1}+1} \]We need \( a_{n+1} < a_n \). If this inequality holds for all \( n \), the terms are decreasing.

Step 4: Testing Decreasing Nature

To verify \( a_{n+1} < a_n \), consider the inequality:\[ \frac{3 \sqrt{n+2}}{\sqrt{n+1} + 1} < \frac{3 \sqrt{n+1}}{\sqrt{n} + 1} \]Instead, we simplify checking a simpler version of this or find the derivative of a lengthier form of \( a_n \) to establish decreasing behavior. Alternatively, numerically testing shows that \( a_n \) does not always decrease consistently. Therefore, we proceed to test the limits.

Step 5: Checking the Limit Condition

Check if \( \lim_{n \to \infty} a_n = 0 \):\[\lim_{n \to \infty} \frac{3 \sqrt{n+1}}{\sqrt{n}+1} = \lim_{n \to \infty} \frac{3(\sqrt{n(1+1/n)})}{\sqrt{n}(1+1/n^{1/2})} \approx \lim_{n \to \infty} 3 \frac{\sqrt{1+1/n}}{1+1/\sqrt{n}} \]As \( n \to \infty \), both terms in the fraction approach 1, so the limit is 3, not 0. Hence, \( a_n \) does not meet the \( \lim_{n \to \infty} a_n = 0 \) requirement for convergence.

Key Concepts

Alternating Series Test

When we talk about alternating series, we are dealing with sums where the signs of the terms alternate back and forth between positive and negative. A standard test used in mathematical analysis to determine if such series converge is the Alternating Series Test (AST). This involves checking a few conditions:

• The series must alternate signs. By looking at our series, we see the term \((-1)^{n+1}\) ensuring this alternation.
• The absolute value of the terms, \(a_n\), should decrease as \(n\) increases.
• Finally, the sequence of absolute values, \(a_n\), should tend toward zero as \(n\) approaches infinity.

Meeting these conditions ensures convergence. However, if any of these conditions fail, we may not conclude convergence solely using the AST.

Convergence of Series

Series convergence is a core concept in calculus where we determine if a sum of infinitely many terms actually approaches a finite value. For alternating series, if all conditions of the Alternating Series Test are met:

• The terms \(a_n\) are positive,
• The terms \(a_n\) decrease sequentially,
• As \(n\) increases, \(a_n\) tends toward zero,

the series converges. If any condition is unmet, other tests or techniques must be explored for convergence. In our exercise, since the last condition regarding the limit of sequence is not satisfied (since the limit is 3, not 0), the series diverges.

Limit of Sequence

A core aspect of the Alternating Series Test is that each term in the series \(a_n\) should approach zero as \(n\) increases significantly. This can be checked using the limit of the sequence formulation \( \lim_{n \to \infty} a_n = 0 \). This ensures that as we sum an infinite number of terms, the impact of adding additional terms becomes negligible. In our exercise, checking this condition revealed that \( \lim_{n \to \infty} \frac{3 \sqrt{n+1}}{\sqrt{n}+1} \) equates to 3, not 0. This means the terms do not diminish and the series does not converge.

Decreasing Sequence

For the Alternating Series Test, it's crucial for the sequence of absolute terms to decrease. Simply put, each term should be less than or equal to the one that comes before it, making it a decreasing sequence. To verify this, we compare two successive terms \(a_n\) and \(a_{n+1}\):

• \(a_n = \frac{3 \sqrt{n+1}}{\sqrt{n}+1}\)
• \(a_{n+1} = \frac{3 \sqrt{n+2}}{\sqrt{n+1}+1}\)
• To show \(a_{n+1} < a_n\) is to confirm decreasing behavior.

The solution tries to numerically test or uses derivatives to establish this; however, the inequality and numerical tests suggest that \(a_n\) does not consistently decrease. Concluding, we cannot rely on this decreasing condition to argue for convergence using the Alternating Series Test.