Problem 36 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The sequence diverges as it alternates between \(-1\) and \(1\).

Step 1: Analyze the Sequence

Given the sequence \( a_n = \frac{(-1)^{n + 1}n}{n + \sqrt{n}} \), we need to identify its behavior as \( n \to \infty \). The numerator is \((-1)^{n+1}n\), which alternates signs depending on whether \(n\) is odd or even. The denominator \(n + \sqrt{n}\) grows with \(n\).

Step 2: Simplify the Expression

Focus on the highest power terms in both the numerator and denominator. The term \(n\) is the primary component in the numerator, and in the denominator, \(n\) is dominant as \(\sqrt{n}\) grows slower than \(n\). Simplify: \( a_n = \frac{(-1)^{n+1}n}{n+\sqrt{n}} = \frac{(-1)^{n+1}}{1 + \frac{\sqrt{n}}{n}} \).

Step 3: Evaluate the Limit

Consider \(\lim_{n \to \infty} \frac{(-1)^{n+1}}{1 + \frac{\sqrt{n}}{n}}\). Note that \(\frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}}\), which approaches 0 as \(n \to \infty\). Thus, the sequence approaches \(\lim_{n \to \infty} \frac{(-1)^{n+1}}{1} = (-1)^{n+1}\).

Step 4: Check Alternating Behavior

The sequence values alternate between \(-1\) and \(1\) as \((-1)^{n+1}\) switches signs with each consecutive integer. This indicates that the sequence does not settle at a single value.

Step 5: Conclusion

Since the sequence does not approach a single number but alternates between \(-1\) and \(1\), it diverges. The alternating behavior prevents convergence because it lacks a fixed value as \(n \to \infty\).

Key Concepts

Alternating Sequences

When you see expressions like \((-1)^{n+1}\), you are dealing with an alternating sequence. Alternating sequences switch their terms based on whether the index (\(n\)) is odd or even. This causes the sequence to change direction, bouncing back and forth between positive and negative values. In our given sequence \(a_n = \frac{(-1)^{n+1}n}{n + \sqrt{n}}\), the alternating factor \((-1)^{n+1}\) makes \(a_n\) switch signs with every consecutive term. As a result:

• If \(n\) is odd, \(a_n\) is positive.
• If \(n\) is even, \(a_n\) is negative.

This alternating behavior is crucial because it influences whether the sequence might converge, that is, approach a single value as \(n\) goes to infinity. Other components of the sequence further complicate this, determining if the oscillation reduces over time.

Convergence and Divergence

Convergence and divergence are central in understanding sequences. A sequence is said to converge if, as you go further along the sequence (i.e., as \(n\) approaches infinity), the terms get closer and closer to a specific number, known as the limit. In contrast, a sequence diverges if its terms do not settle towards any particular value.Let's revisit our sequence: \(a_n = \frac{(-1)^{n+1}n}{n + \sqrt{n}}\). The simplifying step gives us \(a_n = \frac{(-1)^{n+1}}{1 + \frac{\sqrt{n}}{n}}\). As \(n\) approaches infinity, \(\frac{\sqrt{n}}{n}\) approaches 0, suggesting that the effect of \(\sqrt{n}\) diminishes in comparison to \(n\). This simplification helps understand the limiting behavior.Since the numerator still alternates between positive and negative ones, the sequence doesn't settle down to a single value. Instead, it keeps bouncing between -1 and 1, leading to the conclusion that the sequence diverges. It embodies both convergence's counterpart and the fascinating world of divergent sequences.

Limits of Sequences

The limit of a sequence is a notion that describes where the sequence is "heading" as \(n\) becomes infinitively large. When evaluating a limit, we examine the behavior of the sequence's terms when taken to the extreme.For the sequence \(a_n = \frac{(-1)^{n+1}n}{n + \sqrt{n}}\), the main task was to find out whether the sequence converges and what the limit would be if it does. Simplifying, \(a_n = \frac{(-1)^{n+1}}{1 + \frac{\sqrt{n}}{n}}\) gives us valuable insight. As \(n\) grows, \(\frac{\sqrt{n}}{n}\) approaches 0. So, we were guided toward a sequence that behaves like \(\frac{(-1)^{n+1}}{1}\)—simply \( (-1)^{n+1}\). The alternating term means the sequence vacillates indefinitely and does not zero in on a particular number. This confirms our understanding of limits in sequences, underscoring the idea that this sequence doesn't approach a single number, thus diverges.