Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

An explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ _{n}=(-1)^{n} \frac{n}{n+2} $$

Step-by-Step Solution

Answer

The sequence diverges.

Step 1: Identify the Explicit Formula

The given formula for the sequence is \(a_{n} = (-1)^{n} \frac{n}{n+2}\). This formula will be used to find the terms of the sequence \(\{a_{n}\}\).

Step 2: Calculate the First Five Terms

Plug in the values of \(n = 1, 2, 3, 4,\) and \(5\) into the formula to find the first five terms.- \(n = 1\): \(a_{1} = (-1)^{1} \frac{1}{1+2} = -\frac{1}{3}\)- \(n = 2\): \(a_{2} = (-1)^{2} \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}\)- \(n = 3\): \(a_{3} = (-1)^{3} \frac{3}{3+2} = -\frac{3}{5}\)- \(n = 4\): \(a_{4} = (-1)^{4} \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}\)- \(n = 5\): \(a_{5} = (-1)^{5} \frac{5}{5+2} = -\frac{5}{7}\)

Step 3: Determine Sequence Convergence or Divergence

Examine the behavior of the sequence as \(n\) approaches infinity. The explicit formula \(a_n = (-1)^n \frac{n}{n+2}\) suggests that the terms alternate in sign and that the fraction \(\frac{n}{n+2}\) approaches 1 as \(n\) becomes very large. However, due to the alternating factor \((-1)^n\), the sequence does not settle at any particular value, implying it diverges.

Step 4: Conclusion on Limit

Since the sequence \(a_n = (-1)^n \frac{n}{n+2}\) continues to oscillate as \(n\) increases and does not approach a fixed number, the limit \(\lim_{n \rightarrow \infty} a_{n}\) does not exist. Thus, the sequence diverges.

Key Concepts

Alternating Sequences

An alternating sequence involves terms that change signs between positive and negative. This behavior is commonly represented in sequences by a factor of \((-1)^n\). For the sequence given by \(a_{n} = (-1)^{n} \frac{n}{n+2}\), the \((-1)^n\) term ensures that the sequence alternates signs depending on whether \(n\) is even or odd.
Examples of how this works include:

• For \(n = 1\), the term is \(-\frac{1}{3}\), which is negative.
• For \(n = 2\), the term is \(+\frac{1}{2}\), which is positive.
• The pattern continues with every odd \(n\) producing a negative term and every even \(n\) producing a positive term.

The key characteristic here is not just the change of signs, but how these alternating signs interact with the underlying fraction. This interplay affects the sequence's overall behavior as it progresses.

Limit Analysis

The analysis of limits is crucial in determining whether a sequence converges or diverges. When examining the limit of our sequence \(a_{n} = (-1)^{n} \frac{n}{n+2}\), focus on the fraction part \(\frac{n}{n+2}\).
As \(n\) grows larger, the value of \(\frac{n}{n+2}\) approaches \(1\) because the \(+2\) in the denominator becomes negligible compared to \(n\).

The fraction converges to a value of \(1\), but the crucial part of this sequence is the alternating factor \((-1)^n\), which causes each term to toggle between positive and negative. When analyzing limits in such contexts, focus not only on the magnitudes or absolute values of terms but also on their oscillatory behavior.

Divergence

A sequence diverges if it does not approach a definite limit as \(n\) tends to infinity. For our sequence \(a_{n} = (-1)^{n} \frac{n}{n+2}\), the fluctuation between negative and positive values ensures divergence.
This is because, although the absolute magnitude of the sequence's terms gets closer to \(1\) due to the fraction \(\frac{n}{n+2}\) trending towards \(1\), there is no single value that captures the 'settling' point of the sequence.

• The sequence oscillates indefinitely.
• It fails to consistently approach a specific value or range.

Thus, the sequence's lack of convergence confirms its divergence, meaning that \(\lim_{n \to \infty} a_{n}\) does not exist because it continues to fluctuate without stabilizing.