Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^{4}}{2^{n}} $$

Step-by-Step Solution

Answer

The series is absolutely convergent

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^{4}}{2^{n}} \). This is an alternating series because it includes the factor \((-1)^{n+1}\), which causes the sign of the terms to alternate.

Step 2: Apply the Alternating Series Test

The Alternating Series Test states that a series \( \sum (-1)^n a_n \) converges if the sequence \( a_n \) (the absolute value of the terms) is positive, decreasing, and approaches zero. For our series, \( a_n = \frac{n^4}{2^n} \). Check if this satisfies the conditions.

Step 3: Verify \( a_n \to 0 \)

Analyze \( a_n = \frac{n^4}{2^n} \). As \( n \to \infty \), the denominator grows exponentially faster than the polynomial in the numerator, so \( a_n \to 0 \).

Step 4: Check Monotonicity

Next, we must determine if \( a_n = \frac{n^4}{2^n} \) is eventually decreasing. By considering the ratio \( \frac{a_{n+1}}{a_n} = \frac{(n+1)^4}{2^{n+1}} \times \frac{2^n}{n^4} = \frac{(n+1)^4}{2n^4} \), we can show that this is less than 1 for sufficiently large \( n \), confirming that \( a_n \) is decreasing.

Step 5: Investigate Absolute Convergence

An alternating series is absolutely convergent if \( \sum |a_n| \) converges. Here, consider the non-alternating series \( \sum_{n=1}^{\infty} \frac{n^4}{2^n} \). Use the Ratio Test to check its convergence.

Step 6: Apply the Ratio Test for Absolute Convergence

For the series \( \sum \frac{n^4}{2^n} \), calculate the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)^4}{2n^4} \). Simplifying gives \( \frac{1}{2} \) as \( n \to \infty \), since the exponential term dominates. Therefore, because \( L < 1 \), the series absolutely converges.

Key Concepts

Alternating Series Test

The given series \[ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{n^4}{2^n} \] is an alternating series. An alternating series means that its terms switch signs between positive and negative, which is represented by the \((-1)^{n+1}\) term. To determine if such a series converges, we use the Alternating Series Test. This test requires the following conditions:

• The sequence of absolute values, \(a_n\), must be positive.
• The sequence must be decreasing.
• The sequence must tend to zero as \(n\) go to infinity.

In our series, \(a_n = \frac{n^4}{2^n}\). First, check if \(a_n\) approaches zero. Here, the denominator, \(2^n\), grows exponentially and dominates the polynomial \(n^4\), making \(a_n \to 0\). For decreasing behavior, examining \(\frac{a_{n+1}}{a_n}\) reveals it's less than 1 for large \(n\), confirming decrease. Thus, by these reasons, our series converges by the Alternating Series Test.

Absolute Convergence

Absolute convergence provides a deeper insight into the convergence of series. A series is said to be "absolutely convergent" if the series of absolute values converges. For our series, checking absolute convergence involves considering the non-alternating version:\[ \sum_{n=1}^{\infty} \frac{n^4}{2^n} \]If this series converges, then the original alternating series will also absolutely converge.Exploring absolute convergence helps to determine not just whether a series converges, but also the nature of its convergence. When it's shown that \(\sum |a_n|\) converges, it implies both the alternating behavior is irrelevant to this, and the series converges absolutely, which is even stronger and more useful in mathematical analysis.

Ratio Test

The Ratio Test is an essential tool to explore the absolute convergence of series. It calculates the limit \(L\) by taking:\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.\]For our series, \(\sum \frac{n^4}{2^n}\), we calculate:\[ \frac{(n+1)^4}{2n^4} \to \frac{1}{2},\]as \(n\) goes to infinity. Since \(L < 1\), this indicates that the series is absolutely convergent.The Ratio Test is straightforward and effective, especially where exponential functions are involved. If the limit \(L < 1\), then the series converges absolutely. If \(L > 1\), it diverges, and if \(L = 1\), the test is inconclusive, and other methods may be necessary. In this exercise, the Ratio Test affirms that the convergence of our series is indeed absolute, solidifying understanding of the convergence behavior.