Problem 23 - Calculus: Early Transcendentals

Question

Show that the series is convergent. How many terms of the series do we need to add in order to find the sum to the indicated accuracy? \( \displaystyle \sum_{n = 1}^{\infty} \frac{( - 1)^{n+1}}{n^6} (|error| < 0.00005) \)

Step-by-Step Solution

Answer

The series converges, and 6 terms are needed for the desired accuracy.

Step 1: Understand the Series

The given series \( \sum_{n = 1}^{\infty} \frac{(-1)^{n+1}}{n^6} \) is an alternating series because the terms alternate in sign as \( n \) increases.

Step 2: Convergence Criteria

To show that the series converges, we can use the Alternating Series Test. An alternating series \( \sum_{n=1}^{\infty} (-1)^{n+1}a_n \) converges if the sequence \( a_n \) is positive, decreasing, and \( \lim_{n \to \infty} a_n = 0 \). In this series, \( a_n = \frac{1}{n^6} \).

Step 3: Check Positivity

Since \( a_n = \frac{1}{n^6} \) and \( n^6 > 0 \) for all \( n \geq 1 \), the terms \( a_n \) are positive.

Step 4: Check Monotonic Decrease

The sequence \( \frac{1}{n^6} \) is decreasing because for all \( n \geq 1 \), \( \frac{1}{(n+1)^6} < \frac{1}{n^6} \).

Step 5: Limit of a_n

Compute the limit: \( \lim_{n \to \infty} \frac{1}{n^6} = 0 \). Since the limit is zero, the conditions of the Alternating Series Test are met, confirming that the series converges.

Step 6: Determine Number of Terms for Accuracy

For alternating series, the error after \( N \) terms is less than the absolute value of the first omitted term. We need the least \( N \) such that \( \left| \frac{(-1)^{N+2}}{(N+1)^6} \right| < 0.00005 \). Simplifying, this is equivalent to finding \( N \) for \( \frac{1}{(N+1)^6} < 0.00005 \).

Step 7: Solve for N

We solve \( \frac{1}{(N+1)^6} < 0.00005 \) by finding \( (N+1)^6 > \frac{1}{0.00005} = 20000 \). Taking the sixth root, \( N+1 > 20000^{1/6} \approx 6.58 \). Thus, \( N \geq 6 \), so we need \( N = 6 \) terms.

Key Concepts

Convergence Criteria

The first concept to tackle when analyzing whether a series like \( \sum_{n = 1}^{\infty} \frac{(-1)^{n+1}}{n^6} \) converges is understanding the convergence criteria. There are specific tests available in mathematics to assist in identifying if an infinite series converges to a specific value. For alternating series, the Alternating Series Test is particularly helpful.
In this problem, our series goes from \( n = 1 \) to infinity and has terms of the form \( \frac{(-1)^{n+1}}{n^6} \). Therefore, each term alternates between positive and negative. The convergence of the series can be determined if it meets the following criteria:

• The sequence of absolute values \( a_n = \frac{1}{n^6} \) is positive.
• The sequence \( a_n = \frac{1}{n^6} \) decreases in a monotonic manner as \( n \) increases.
• The limit of \( a_n \) as \( n \) approaches infinity is zero, \( \lim_{n \to \infty} a_n = 0 \).

By confirming these conditions, the test confirms that the series in question indeed converges.

Alternating Series Test

The Alternating Series Test is a fundamental tool when it comes to determining the convergence of alternately signed series. This test focuses on the nature of the terms themselves rather than the entire sum.
For a series \( \sum_{n=1}^{\infty} (-1)^{n+1}a_n \) to converge under this test, it's essential that:
The terms \( a_n \) need to be a positive sequence, so meeting \( a_n > 0 \) is crucial. In our specific series, \( a_n = \frac{1}{n^6} \), and as long as \( n \geq 1 \), each term is positive, since \( n^6 \) is always positive.
Next, the sequence must be monotonically decreasing, meaning that each term is less than its predecessor. The sequence \( \frac{1}{n^6} \) naturally decreases as the denominator grows much faster than the numerator remains at 1.
Furthermore, it is vital that the sequence approaches zero as \( n \) heads towards infinity. Calculating \( \lim_{n \to \infty} \frac{1}{n^6} = 0 \), firmly seals the final condition needed for convergence confirmation.

Error Estimation

When dealing with alternating series such as \( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^6} \), knowing how many terms to add in order to reach a certain accuracy is crucial. This is where error estimation comes into play.
For an alternating series, a neat rule can be applied: the absolute error of truncating the series after \( N \) terms is less than the size of the first omitted term. In mathematical terms, the error of stopping at term \( N \) is less than \( |a_{N+1}| \).

• For our series, we need to establish \( \left| \frac{1}{(N+1)^6} \right| < 0.00005 \) to ensure our error is bounded by our tolerance.
• Upon solving \( \frac{1}{(N+1)^6} < 0.00005 \), we find \( (N+1)^6 > 20000 \).
• By taking the sixth root, it becomes apparent \( N+1 > 20000^{1/6} \approx 6.58 \).
• Thus, a minimum \( N \) of 6 terms is necessary to be within the threshold.

This ensures that our calculation achieves the desired accuracy with the confidence that the error will not exceed the predetermined bound.

Positive Decreasing Sequence

A key element of utilizing the Alternating Series Test effectively lies in confirming the sequence \( a_n = \frac{1}{n^6} \) as both positive and decreasing. This condition is fundamental for the convergence of the series to hold true without ambiguity.
The positivity of \( a_n \) is straightforward: since \( n^6 \) is positive for all integers \( n \geq 1 \), naturally \( \frac{1}{n^6} \) remains positive for all such \( n \) values in our series.
Monotonicity, or the property of being decreasing, necessitates that every succeeding term is smaller than the preceding term. Examining the function \( \frac{1}{n^6} \), it is clear that as \( n \) increases, \( n^6 \) grows larger, thus making the fraction \( \frac{1}{n^6} \) shrink smaller. Hence, we see:

• \( \frac{1}{(n+1)^6} < \frac{1}{n^6} \) for all \( n \geq 1 \).

This property ensures that the series \( a_n \) meets the necessary criteria to apply the Alternating Series Test confidently, confirming that it is both positive and decreasing throughout the series. This alignment cements the test's validity and confirms the series converges.