Problem 7 - Essential Calculus Early Transcendentals

Question

\(3-8=\) Test the series for convergence or divergence. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{3 n-1}{2 n+1} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify Test for Alternating Series

To determine the convergence of the series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{3n-1}{2n+1} \), notice that it is an alternating series. We can apply the Alternating Series Test, which states that the series \( \sum (-1)^{n} a_n \) converges if \( a_n \) is decreasing and \( \lim_{n \to \infty} a_n = 0 \).

Step 2: Define the Terms of the Series

The terms \( a_n = \frac{3n-1}{2n+1} \) are what we will test using the conditions of the Alternating Series Test. We need to check if these terms are decreasing and if their limit is zero as \( n \to \infty \).

Step 3: Test Monotonicity

To show that \( a_n = \frac{3n-1}{2n+1} \) is decreasing, check the derivative \( a'_n \) or compare \( a_{n+1} \) and \( a_n \). If \( a_{n+1} < a_n \), then the sequence is decreasing. Calculate:\[ a_{n+1} = \frac{3(n+1)-1}{2(n+1)+1} = \frac{3n+2}{2n+3} \]Compare:\[ a_{n+1} < a_n \iff \frac{3n+2}{2n+3} < \frac{3n-1}{2n+1} \]

Step 4: Simplify the Inequality from Step 3

Cross-multiply and simplify the inequality from Step 3:\[ (3n+2)(2n+1) < (3n-1)(2n+3) \]Expand both sides:\[ 6n^2 + 2n + 3n + 2 < 6n^2 + 9n - 2n - 3 \]This simplifies to:\[ 5n + 2 < 7n - 3 \]\[ 5n + 2 < 7n - 3 \iff 2 < 2n - 3 \iff 5 < 2n \iff n > 2.5 \]Since \( n \) is an integer, the inequality holds for all \( n \geq 3 \). Thus, \( a_n \) is decreasing for \( n \geq 3 \).

Step 5: Check the Limit Condition

Next, check \( \lim_{n \to \infty} \frac{3n-1}{2n+1} \):Divide the numerator and denominator by \( n \):\[ \lim_{n \to \infty} \frac{3n-1}{2n+1} = \lim_{n \to \infty} \frac{3 - \frac{1}{n}}{2 + \frac{1}{n}} = \frac{3}{2} \]The limit is not zero. Therefore, the alternating series does not meet the condition \( \lim_{n \to \infty} a_n = 0 \) of the Alternating Series Test.

Key Concepts

series convergence

Understanding series convergence is crucial when analyzing sequences and series in mathematics. A series is simply the sum of the terms of a sequence, and it can either converge or diverge. Convergence implies that as you add more terms, the total approaches a specific value. Divergence, on the other hand, means that as you continue summing the terms, the total grows indefinitely or fluctuates without settling on a value.

When determining convergence, mathematicians use various tests, one of which is the Alternating Series Test. Generally, if a series satisfies the convergence criteria of such tests, it is considered convergent. However, if no test confirms convergence, the series might be divergent. A good starting point is trying to find and apply the most suitable test. This can include simple observations about the behavior of the terms, like whether they decrease in magnitude or approach zero.

alternating series

Alternating series are special types of series where the terms alternate in sign, like \ \( (-1)^n \) multiplied by the term of the sequence. This creates a series that oscillates, moving back and forth across a middle value. A common characteristic of these series is that they frequently converge, even if the terms themselves do not approach zero.

To determine if an alternating series converges, we use the Alternating Series Test. Here are the key conditions for convergence under this test:

• The absolute value of the terms \( a_n \) must decrease as \( n \) increases after a certain point.
• The limit of the terms \( a_n \) should be zero as \( n \) approaches infinity.

In the exercise given, we are applying this test to a specific alternating series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{3n-1}{2n+1} \). While the terms \( \frac{3n-1}{2n+1} \) are decreasing, they do not tend to zero, leading to the conclusion that this particular series does not converge by this test.

limit of a sequence

The concept of the limit of a sequence is foundational in understanding both calculus and series analysis. The limit describes the value that the terms of a sequence \( a_n \) approach as the index \( n \) becomes very large. If the terms of a sequence settle on a specific number as \( n \) goes to infinity, then the sequence is said to "converge" to this limit. Otherwise, it diverges.

In the context of the Alternating Series Test, the limit of a sequence plays a crucial role in checking convergence. The test specifically requires that \( a_n \rightarrow 0 \). If this condition fails, then there's strong evidence that the series does not converge. For the example in our exercise, although the terms do not settle to zero but instead approach \( \frac{3}{2} \), the convergence condition regarding the limit is not satisfied, therefore implying divergence according to the Alternating Series Test.