Problem 31 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{n}{n+1} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Check Absolute Convergence

To check for absolute convergence, consider the series of absolute values: \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{n}{n+1} \right| = \sum_{n=1}^{\infty} \frac{n}{n+1} \).The terms \( \frac{n}{n+1} \) approach 1 as \( n \to \infty \), which do not tend to zero. According to the Divergence Test, if the limit of the terms does not approach zero, the series diverges. Therefore, \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{n}{n+1} \right| \) diverges, so the original series does not converge absolutely.

Step 2: Check Conditional Convergence

We check if the original series \( \sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} \) converges. The series has alternating terms, so we apply the Alternating Series Test.The test requires:1. The sequence \( a_n = \frac{n}{n+1} \) must decrease.2. The limit \( \lim_{n \to \infty} a_n = 0 \).The sequence \( a_n \) does not go to zero as \( n \to \infty \) since \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \), which violates the second condition of the Alternating Series Test. Therefore, the series does not converge conditionally.

Step 3: Conclusion on Convergence and Divergence

Since the series \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{n}{n+1} \right| \) does not converge, it does not converge absolutely. Additionally, the original series \( \sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} \) does not converge conditionally due to failing the Alternating Series Test. Therefore, the series diverges.

Key Concepts

Absolute Convergence

Absolute convergence is a way to determine if a series will sum up to a finite value even when taking all terms as positive. To test for absolute convergence, replace each term in the series with its absolute value and check if this new series converges. Essentially, it's like asking: "Does the series converge if all its terms are positive?"
To illustrate, instead of dealing with the original series \( \sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} \), we consider \( \sum_{n=1}^{\infty} \left| (-1)^n \frac{n}{n+1} \right| = \sum_{n=1}^{\infty} \frac{n}{n+1} \).

• If this new series converges, the series converges absolutely.
• If it diverges, the series does not have absolute convergence.

In our case, since the terms diverge as \( n \to \infty \), we conclude that the series \( \sum_{n=1}^{\infty} \frac{n}{n+1} \) diverges. Therefore, the original series does not converge absolutely.

Alternating Series Test

The Alternating Series Test is useful for determining the convergence of series whose terms alternate in sign. This means the terms go back and forth between positive and negative. The series \( \sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} \) is such a series. The test involves two conditions:

• The terms \( a_n = \frac{n}{n+1} \) must decrease in value with each increase in \( n \).
• The limit of \( a_n \) as \( n \to \infty \) must be zero: \( \lim_{n \to \infty} a_n = 0 \).

If both conditions are met, the series converges. In our example:
- The terms do not decrease monotonically since the terms approach 1 as \( n \to \infty \).- The second condition fails as well because \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \), which is not zero.
Consequently, this series does not converge conditionally.

Divergence Test

The Divergence Test, also known as the nth-term test for divergence, is a quick check to see if a series can possibly converge. It states that if the limit of the terms of a series as \( n \to \infty \) is not zero, then the series must diverge.

• This doesn't tell us if a series converges, only that it can't if the test fails.
• For the series \( \sum_{n=1}^{\infty} (-1)^n \frac{n}{n+1} \), we found that \( \lim_{n \to \infty} \frac{n}{n+1} = 1 \).

Since the limit of the terms is not zero, by the Divergence Test, we can confidently say that this series diverges.
This test was crucial as it immediately ruled out absolute convergence once the absolute values of the terms were considered.