Problem 51 - Thomas Calculus

Question

In Exercises \(49 - 52 ,\) estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. $$ \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n + 1 } \frac { ( 0.01 ) ^ { n } } { n } $$ As you will see in Section 10.7 the sum is \(\ln ( 1.01 ) .\)

Step-by-Step Solution

Answer

The error magnitude using four terms is less than \(2 \times 10^{-11}\).

Step 1: Identify the Series Type

Given series is an alternating series: \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(0.01)^n}{n} \). Recognize this as an alternating series due to the \((-1)^{n+1}\) term.

Step 2: Determine the Alternating Series Approximation Error

For an alternating series \( \sum (-1)^n a_n \), the error in approximating the sum by the first \(N\) terms is less than or equal to the first neglected term: \(|R_N| = |a_{N+1}|\).

Step 3: Calculate the First Neglected Term

Here, \(a_n = \frac{(0.01)^n}{n}\). To find the error for the first four terms, compute the fifth term: \(a_5 = \frac{(0.01)^5}{5} = \frac{10^{-10}}{5} = 2\times10^{-11} \).

Step 4: Interpret the Result

The magnitude of the error in approximating the sum of the entire series using the first four terms is less than \(2 \times 10^{-11}\). This indicates a highly accurate approximation.

Key Concepts

Error Estimation

Error estimation in the context of series is crucial to determine how close an approximation is to the actual sum of an infinite series. When dealing with alternating series, which have terms that switch signs, estimating the error becomes more straightforward. This is because for a convergent alternating series, the error in the approximation using the first few terms can be bounded easily. For example, when we approximate an alternating series using its first four terms, the error is determined by the magnitude of the fifth term. In mathematical terms, if the series is \( \sum_{n=1}^{\infty} (-1)^{n+1} a_n\), then the error in approximating it using the first four terms \( R_4 \) is given by the absolute value of the next term, \( |a_5| \). This tells us that the error won't exceed the size of this next term. In our specific problem, the error for the series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(0.01)^n}{n} \) when estimated using the first four terms is less than \(2 \times 10^{-11}\), making it a very small and thus highly accurate approximation.

Series Approximation

Approximating a series involves using a finite number of terms to represent an infinite sum. This is especially important in instances where the actual sum of the series is difficult or impossible to compute exactly. In the case of alternating series, we focus on using a few terms to achieve an effective approximation with minimal error. An approximation \( S_N \) of an infinite series \( S = \sum_{n=1}^{\infty} a_n \) can be calculated by summing the first \( N \) terms: \( S_N = \sum_{n=1}^{N} a_n \). The quality of this approximation depends on how quickly the terms \( a_n \) decrease in size. Faster decrease means a better approximation with fewer terms. In our case, the series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(0.01)^n}{n} \) is approximated by summing the first four terms. Since each term quickly becomes tiny given that the factor \( (0.01)^n \) reduces exponentially, our approximation is notably accurate, as evidenced by the very small error after four terms.

Alternating Series Test

The alternating series test is a method used to determine the convergence of a particular type of series known as alternating series. An alternating series is characterized by terms that alternate in sign, such as \( \sum_{n=1}^{\infty} (-1)^n a_n \). The test provides conditions under which these series converge. For an alternating series to converge, the terms \( a_n \) must be decreasing in magnitude, and their limit must approach zero as \( n \) tends to infinity. This simple yet powerful test helps us understand and predict the behavior of alternating series. In our example, the alternating series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(0.01)^n}{n} \) is convergent because the terms \( \frac{(0.01)^n}{n} \) are positive and decrease to zero. The alternating series test thus assures us that the series converges, which supports the validity of our approximation and error estimation. Knowing the series converges gives confidence in using this approach for practical calculations.