Problem 6 - Active Calculus

Question

We have shown that if \(\sum(-1)^{k+1} a_{k}\) is a convergent alternating series, then the sum \(S\) of the series lies between any two consecutive partial sums \(S_{n}\). This suggests that the average \(\frac{S_{n}+S_{n+1}}{2}\) is a better approximation to \(S\) than is \(S_{n}\). a. Show that \(\frac{S_{n}+S_{n+1}}{2}=S_{n}+\frac{1}{2}(-1)^{n+2} a_{n+1}\). b. Use this revised approximation in (a) with \(n=20\) to approximate \(\ln (2)\) given that $$ \ln (2)=\sum_{k=1}^{\infty}(-1)^{k+1} \frac{1}{k} . $$ Compare this to the approximation using just \(S_{20} .\) For your convenience, \(S_{20}=\frac{155685007}{232792560}\).

Step-by-Step Solution

Answer

Using the revised approximation gives a closer value \( \frac{161227687}{232792560} \) compared to \( S_{20} = \frac{155685007}{232792560} \).

Step 1 - Establish the Series Definition and Partial Sums

Define the alternating series sum and its partial sums: \[ S_n = \sum_{k=1}^{n} (-1)^{k+1} a_k \] and \[ S_{n+1} = \sum_{k=1}^{n+1} (-1)^{k+1} a_k = S_n + (-1)^{n+2} a_{n+1} \]

Step 2 - Average of Consecutive Partial Sums

Write the average of two consecutive partial sums: \[ \frac{S_n + S_{n+1}}{2} \] Substitute the expression for \(S_{n+1}\): \[ \frac{S_n + (S_n + (-1)^{n+2} a_{n+1})}{2} \]

Step 3 - Simplify the Expression

Simplify the expression for the average: \[ = \frac{2S_n + (-1)^{n+2} a_{n+1}}{2} = S_n + \frac{1}{2}(-1)^{n+2} a_{n+1} \] This confirms the required equation: \[ \frac{S_n + S_{n+1}}{2} = S_n + \frac{1}{2}(-1)^{n+2} a_{n+1} \]

Step 4 - Special Case: Logarithm Series

Given the series \[ \ln(2) = \sum_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} \] with \( a_k = \frac{1}{k} \), calculate the approximation for \(n=20\).

Step 5 - Applying the Result

Using \( n = 20 \), \[ S_{20} = \frac{155685007}{232792560} \]. Compute \[ \frac{S_{20} + S_{21}}{2} = S_{20} + \frac{1}{2}(-1)^{22} a_{21} \] Since \( (-1)^{22} = 1 \) and \( a_{21} = \frac{1}{21} \), the result is: \[ \frac{155685007}{232792560} + \frac{1}{2 \cdot 21} = \frac{155685007}{232792560} + \frac{1}{42} \]

Step 6 - Evaluate and Compare

Evaluate \( \frac{1}{42} = \frac{232792560}{42} = 5542680 \), so \[ S_{20} + \frac{1}{42} = \frac{155685007 + 5542680}{232792560} = \frac{161227687}{232792560} \] Compare this to \( S_{20} = \frac{155685007}{232792560} \). The approximation \( \frac{161227687}{232792560} \) is closer to \( \ln(2) \) than \( \frac{155685007}{232792560} \).

Key Concepts

Convergent Series

A convergent series is a series where the sum of its terms approaches a specific value as more terms are added. In mathematical terms, a series \(\[ \sum_{k=1}^{\text{∞}} a_k \]\) is said to be convergent if its sequence of partial sums converges to a limit. This means that as you add up the terms one by one, the total gets closer and closer to this limit.

Alternating series, like the one in our problem, often show this behavior. For an alternating series to be convergent, the absolute value of the terms must decrease over time and approach zero. This is often confirmed using the Alternating Series Test.

Identifying whether a series is convergent is crucial. If a series is convergent, its sum exists and can be approximated using various methods, as we will see.

Partial Sum Approximation

When dealing with endless series, calculating the exact sum is often impossible. Instead, we rely on partial sums to approximate the total sum. A partial sum is simply the sum of the first n terms of a series. We denote it as \( S_n = \sum_{k=1}^{n} a_k \) where n is the number of terms included.

The interesting property of alternating series is that the sum of the series \( S \) lies between any two consecutive partial sums, \( S_n \) and \( S_{n+1} \). By averaging these two partial sums, \( \frac{S_n + S_{n+1}}{2} \), we obtain a better approximation of the true series sum. This is because the oscillation in the values is smoothed out, providing a closer estimate to the actual limit.

Natural Logarithm Approximation

One fascinating application of series summation is approximating functions like the natural logarithm. In the given problem, we use an alternating series to approximate \( \ln(2) \). The series representing this is: \( \ln(2) = \sum_{k=1}^{\text{∞}} (-1)^{k+1} \frac{1}{k} \).

This alternating series converges because its terms decrease in absolute value and approach zero. By calculating partial sums up to a specific term and applying our averaging method, we get an approximation of \( \ln(2) \) that becomes more accurate the further we go. For instance, in the problem, we use up to the 20th term (\( S_{20} \)) and notice that the revised approximation provides a value closer to \( \ln(2) \) than using just the 20th partial sum.

Series Summation

Summation of series—especially infinite series—can be daunting. The key lies in understanding the properties of the series and applying appropriate techniques to approximate sums.

In practice, we often work with partial sums. For example, given an infinite series \( \sum_{k=1}^{\text{∞}} a_k \), the nth partial sum \( S_n \) is the sum of the first n terms: \( S_n = \sum_{k=1}^{n} a_k \). To approximate the entire series, we examine these partial sums as n increases.

Alternating series pose unique characteristics where the exact sum lies between consecutive partial sums. This enables the use of averaging to improve estimations. By computing \( \frac{S_n + S_{n+1}}{2} \), we smooth out the oscillations between sums, achieving a superior approximation to the sum of an alternating series like the approximation of \( \ln(2) \) shown in the worked example.