Problem 49

Question

Estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n} $$

Step-by-Step Solution

Verified
Answer
The error is less than 0.2.
1Step 1: Identify the Series Type
The given series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} \) is an alternating series. It follows the form \( \sum_{n=1}^{\infty} (-1)^{n+1} a_n \), where \( a_n = \frac{1}{n} \). According to the Alternating Series Test, this series converges.
2Step 2: Use Alternating Series Estimation Theorem
The Alternating Series Estimation Theorem states that the absolute error \( |R_N| \) in using the sum of the first \( N \) terms to approximate the sum of an alternating series is less than or equal to the absolute value of the first omitted term.
3Step 3: Calculate the First Omitted Term
Since we are using the first four terms to approximate the series, calculate the fifth term \( a_5 = \frac{1}{5} \).
4Step 4: Estimate the Error
The error \( |R_4| \) is bounded by \( |a_5| = \frac{1}{5} = 0.2 \). This means the magnitude of the error when using the first four terms to approximate the series is less than 0.2.

Key Concepts

Error EstimationConvergence of SeriesAlternating Series Test
Error Estimation
In the study of infinite series, especially alternating series, estimating the error when using a finite number of terms to approximate the series is crucial. An alternating series switches signs between consecutive terms. Therefore, error estimation becomes an essential tool to understand how close our finite approximation is to the actual infinite sum.

In this context, the Alternating Series Estimation Theorem provides a straightforward way to estimate this error. It states that the error, denoted as \( |R_N| \), in approximating the sum of an alternating series by its first \( N \) terms, is less than or equal to the magnitude of the first term not included in the partial sum. This non-included term is your error bound.\[|R_N| \leq |a_{N+1}|\]This result is very useful because it gives assurance that our estimate cannot deviate by more than a calculable amount. In our specific problem, using the first four terms to approximate the series means the error is bounded by the fifth term \( a_5 = \frac{1}{5} \), which translates to an error margin of 0.2.
Convergence of Series
The concept of convergence tells us whether an infinite series adds up to a finite value. For alternating series like the one given in the exercise, the convergence is indicated by the fact that we have terms decreasing in absolute value and that the limit of their absolute values approaches zero as \( n \) approaches infinity.

The series in the exercise is an alternating series because consecutive terms have differing signs. Particularly, it is the harmonic series altered to take on alternating signs, represented by:\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}\]Each term's magnitude \( \frac{1}{n} \) decreases as \( n \) increases, meeting the criteria for convergence per the Alternating Series Test. Hence, we know such series converge to a specific value, though calculating this value exactly might be arduous without advanced tools.
Alternating Series Test
The Alternating Series Test gives us a way to determine whether certain series converge. For a series to pass this test, it must satisfy two conditions:
  • The absolute values of the terms \( a_n \) must decrease to zero.
  • The limit of \( a_n \) as \( n \) approaches infinity must be zero.
When these conditions are met, the series converges.

In our problem, the series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} \) meets both these criteria. The terms \( \frac{1}{n} \) indeed decrease and their limit is zero. Thus, the Alternating Series Test confirms convergence. Understanding this test is crucial because it assures both the mathematical feasibility and practicality when working with infinitely many terms. It's also why we use only a few initial terms from the series for practical applications, using error estimation to gauge the precision.