Problem 35 - Calculus: Early Transcendentals

Question

Determine whether the sequence converges or diverges. If it converges, find the limit. \( a_n = \frac {(-1)^n}{2 \sqrt n} \)

Step-by-Step Solution

Answer

The sequence converges to 0.

Step 1: Identify the Sequence

The sequence given is \( a_n = \frac{(-1)^n}{2\sqrt{n}} \). It is an alternating sequence because of the factor \((-1)^n\), which makes every other term positive or negative.

Step 2: Determine the Behavior of the Magnitude

Ignoring the sign change from \((-1)^n\), we consider the magnitude of the terms: \( \left|a_n\right| = \frac{1}{2\sqrt{n}} \). As \( n \to \infty \), \( \sqrt{n} \to \infty \), so \( \frac{1}{2\sqrt{n}} \to 0 \). Thus, the magnitude of the terms approaches 0.

Step 3: Check for Alternating Sequence Convergence

For an alternating sequence \( a_n = (-1)^n b_n \) to converge, \( b_n \) must converge to 0. Here, \( b_n = \frac{1}{2\sqrt{n}} \) converges to 0 as shown in Step 2.

Step 4: Conclude on Convergence

Since \( \left|a_n\right| = \frac{1}{2\sqrt{n}} \to 0 \), and the sequence is alternating, \( a_n \) converges to 0.

Key Concepts

Alternating Sequence

An alternating sequence is a type of sequence where the signs of the terms change between positive and negative as you move along the sequence. This happens because of the \((-1)^n\) factor. In such sequences, odd terms are positive, and even terms are negative, or vice versa. This flip in sign makes alternating sequences interesting and slightly tricky to analyze. In our exercise, the sequence \(a_n = \frac{(-1)^n}{2\sqrt{n}}\) is clearly alternating, as the \((-1)^n\) factor switches the sign of each consecutive term.

Limit of a Sequence

The limit of a sequence refers to the value that the terms of the sequence approach as the sequence progresses towards infinity. To find out the limit of a sequence, we often look at the behavior of the sequence as \(n\ o \infty\). For the sequence in our problem, to find the limit, we focus on what happens to the terms without their alternating signs, i.e., \(|a_n| = \frac{1}{2\sqrt{n}}\). As \(n\) becomes very large, \(\sqrt{n}\) also increases, pushing \(\frac{1}{2\sqrt{n}}\) closer to 0. Hence, the sequence's terms get closer and closer to 0, indicating that the sequence converges to a limit of 0.

Convergence Criteria

For any sequence to converge, it must draw closer to a certain value as \(n\ o \infty\). For alternating sequences, like in our example sequence where \(a_n = (-1)^n b_n\), the convergence is determined when the absolute value of the terms, \(b_n\), approaches 0. If \(b_n = \frac{1}{2\sqrt{n}}\) converges to 0, then the entire sequence \(a_n\) converges. Therefore, in alternating sequences, once the absolute value of the terms moves towards zero as \(n\) increases, the sequence is said to converge.

Sequence Behavior

The behavior of a sequence is crucial in understanding its long-term progression and determining whether it converges to a limit or diverges. In our sequence \(a_n = \frac{(-1)^n}{2\sqrt{n}}\), two aspects dictate its behavior:

• **Alternating Signs**: The \((-1)^n\) results in a sequence of alternating signs, where consecutive terms flip between positive and negative.
• **Decreasing Magnitude**: The factor \(\frac{1}{2\sqrt{n}}\) causes the magnitude of each term to decrease continuously as \(n\) increases.

These two traits of the sequence ensure that it both alternates in sign and pulls its magnitude closer to 0, allowing us to deduce that the sequence converges to zero.