Problem 3 - 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. $$ \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 1 } { n 3 ^ { n } } $$

Step-by-Step Solution

Answer

The series converges as it satisfies the conditions of the Alternating Series Test.

Step 1: Identify the Series

The given series is \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 1 } { n 3 ^ { n } } \). This is an alternating series because of the term \(( - 1 ) ^ { n + 1 }\), which causes the series to alternate between positive and negative terms.

Step 2: Examine the Alternating Series Test

For an alternating series \(\sum (-1)^{n+1} a_n\) to converge, the sequence \(a_n = \frac { 1 } { n 3 ^ { n } }\) must satisfy the following conditions: (1) \(a_n > 0\); (2) \(a_{n+1} \leq a_n\) for all \(n\), or \(a_n\) is eventually non-increasing; and (3) \(\lim_{n \to \infty} a_n = 0\). Let's check these conditions.

Step 3: Check Condition 1: Positivity

The term \(a_n = \frac { 1 } { n 3 ^ { n } }\) is positive for all \( n \geq 1 \) since both the numerator (1) and the denominator (a product of positive terms) are positive. Thus, \(a_n > 0\) for all \(n\).

Step 4: Check Condition 2: Monotonicity

We need to show that \(a_{n+1} \leq a_n\). Calculate \(a_{n+1} = \frac{1}{(n+1)3^{n+1}}\) and compare it to \(a_n = \frac{1}{n3^n}\). We find \(a_{n+1}/a_n = \frac{n}{(n+1)3}\), which is less than or equal to 1 for all \(n \geq 1\). Therefore, \(a_n\) is non-increasing.

Step 5: Check Condition 3: Limit

Evaluate \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac { 1 } { n 3 ^ { n } }\). As \(n\) grows larger, both \(n\) and \(3^n\) increase, and \(3^n\) grows much faster than \(n\). Thus, \(\lim_{n \to \infty} \frac { 1 } { n 3 ^ { n } } = 0\).

Step 6: Conclusion

Since all the conditions of the Alternating Series Test are satisfied, the series \( \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { 1 } { n 3 ^ { n } } \) converges.

Key Concepts

Convergence of Series

When we discuss the convergence of series, we're talking about whether the sum of an infinite series approaches a specific value. The convergence of an alternating series, like the one we're considering here, depends on certain conditions. The Alternating Series Test is crucial for determining convergence. For a series \( \sum (-1)^{n+1} a_n\), where \(a_n\) is positive:

• The sequence \(a_n\) should be positive: \(a_n > 0\) for all \(n\).
• The sequence \(a_n\) should be non-increasing: this means \(a_{n+1} \leq a_n\).
• The limit of \(a_n\) as \(n\) approaches infinity should be zero: \(\lim_{{n \to \infty}} a_n = 0\).

If all these conditions are satisfied, the series converges. In simpler terms, as you keep adding the terms up, the total settles down and doesn't keep growing or oscillating wildly.

Monotonic Sequence

A monotonic sequence is a sequence that is only non-increasing or non-decreasing. In our context, we're interested in non-increasing sequences. This means each term is less than or equal to the previous term. This property is significant because it reassures us that the terms of the sequence \(a_n\) are not getting larger as \(n\) increases but either decreasing or staying the same.

For our specific example, we showed that \(a_n = \frac{1}{n 3^{n}}\) satisfies this condition. After computing \(a_{n+1}/a_n = \frac{n}{(n+1)3}\), which is less than or equal to 1, we confirm that \(a_n\) forms a monotonic sequence. This provides one piece of evidence that supports convergence according to the Alternating Series Test.

Limits of Sequences

The limit of a sequence is what the terms tend to as they go towards infinity. Understanding limits is pivotal to determining convergence. If the terms of the sequence \(a_n\) vanish (approach zero) as \(n\) becomes very large, it satisfies one of the core requirements of the Alternating Series Test.

For instance, consider \(a_n = \frac{1}{n 3^{n}}\). As \(n\) grows larger, both \(n\) and \(3^n\) increase, with \(3^n\) increasing much more rapidly. Hence, the terms become smaller and smaller, effectively approaching zero. Formally expressed, \(\lim_{{n \to \infty}} a_n = 0\). This result is essential because it ensures that the series is approaching a stable sum rather than diverging.