Question

The following alternating series converge to given multiples of \(\pi .\) Find the value of \(N\) predicted by the remainder estimate such that the Nth partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum \(N\) for which the error bound holds, and give the desired approximate value in each case. Up to 15 decimals places, \(\pi=3.141592653589793 .\)[T] The Euler transform rewrites \(S=\sum_{n=0}^{\infty}(-1)^{n} b_{n}\) as \(S=\sum_{n=0}^{\infty}(-1)^{n} 2^{-n-1} \sum_{m=0}^{n}\left(\begin{array}{c}n \\ m\end{array}\right) b_{n-m}\). For the alternating harmonic series, it takes the form \(\ln (2)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}=\sum_{n=1}^{\infty} \frac{1}{n 2^{n}} .\) Compute partial sums of \(\sum_{n=1}^{\infty} \frac{1}{n 2^{n}}\) until they approximate \(\ln (2)\) accurate to within \(0.0001\). How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate \(\ln (2)\).

Short answer

The Euler transform series needs 8 terms for the desired accuracy; the alternating harmonic series needs 5000 terms.

Step-by-step solution

Understand the Series and Error Bound

We are given the series \( \ln(2) = \sum_{n=1}^{\infty} \frac{1}{n 2^{n}} \). We need to compute partial sums of this series until it approximates \( \ln(2) \) to within an error of 0.0001.

Remainder Estimation for Alternating Series

For alternating series, the error in approximating the sum by the nth partial sum is less than or equal to the absolute value of the first omitted term. We calculate this as \( R_N = \frac{1}{(N+1) 2^{N+1}} \).

Determine N for Desired Accuracy

We want \( R_N < 0.0001 \), so we solve the inequality \( \frac{1}{(N+1) 2^{N+1}} < 0.0001 \). By testing values, it can be determined that \( N = 8 \) satisfies this condition.

Calculate Partial Sum with Minimum N

We compute the partial sum \( S_8 = \sum_{n=1}^{8} \frac{1}{n 2^{n}} \). Calculating each term and summing them, \( S_8 = 0.693359375 \).

Compare with Alternating Harmonic Series Approximation

For the alternating harmonic series \( \ln(2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \), the error for the Nth partial sum is \( |\text{error}| < \frac{1}{N+1} \). Testing values, \( N = 5000 \) is needed for the error to be less than 0.0001.

Key concepts

Error Bound

When working with alternating series, the error bound helps us ensure that our approximation is accurate within a specific range. The error in using an Nth partial sum to approximate the sum of an alternating series is less than or equal to the absolute value of the first term omitted in the partial sum.

This is summarized by the formula for the remainder estimate, which is given as:

• For an alternating series: \[ R_N \leq |a_{N+1}| \] where \( R_N \) is the remainder after \( N \) terms.

This tells us that the error is smaller than the next term that would be added if the sum continued. In practice, this allows us to calculate how many terms are needed to reach a desired level of precision. For example, to ensure a precision of 0.0001, we solve the inequality \( R_N < 0.0001 \) using the remainder formula to find the smallest \( N \) where this holds true.

Partial Sums

Partial sums are a way to approximate an infinite series by taking a finite number of terms. If we're dealing with a series \( S = \sum_{n=1}^{\infty} a_n \), a partial sum is the sum of the first \( N \) terms:

• Partial sum \( S_N = a_1 + a_2 + \ldots + a_N \)

As \( N \) increases, \( S_N \) approaches \( S \) more closely. Calculating partial sums is a common approach to see how close we come to a converged value with a certain number of terms.

For example, in the exercise above, the task is to compute partial sums of the series for \( \ln(2) = \sum_{n=1}^{\infty} \frac{1}{n 2^n} \) and to find the number of terms needed to achieve an error less than 0.0001. By calculating \( S_8 \), which is the sum of the first 8 terms, it has been seen that the approximation is precise enough.

Harmonic Series

The harmonic series is a classic infinite series given by:

• \( \sum_{n=1}^{\infty} \frac{1}{n} \)

However, it does not converge. The alternating harmonic series, on the other hand, is convergent. It is used to approximate log(2):

• \( \ln(2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \)

This series converges more quickly when compared to the normal harmonic series because the terms decrease in value and alternate signs.

In the exercise, the alternating harmonic series is compared to \( \sum_{n=1}^{\infty} \frac{1}{n 2^n} \), a related series, to determine which requires fewer terms to reach within the error bound of 0.0001. It turns out the alternating series requires 5000 terms, showing the impact of series characteristics on convergence speed.

Euler Transform

The Euler transform is a technique employed to speed up the convergence of alternating series. It effectively transforms a series into another form that converges faster. This is particularly useful when you want a more efficient computation for sums of alternating series.

For a given series:

• \( S = \sum_{n=0}^{\infty}(-1)^n b_n \)

The Euler transform rewrites this as:

• \( S = \sum_{n=0}^{\infty}(-1)^n 2^{-n-1} \sum_{m=0}^{n} \binom{n}{m} b_{n-m} \)

For example, it changes the alternating harmonic series \( \ln(2) \) to the series \( \ln(2) = \sum_{n=1}^{\infty} \frac{1}{n 2^n} \). This transformed series reaches an accurate approximation faster, needing fewer terms than the original alternating harmonic series. In the exercise case, only 8 terms are needed after transformation, compared to 5000 terms without it, showcasing the power of the Euler transform.