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] \(\frac{\pi}{\sqrt{12}}=\sum_{k=0}^{\infty} \frac{(-3)^{-k}}{2 k+1}\), error \(<0.0001\)

Short answer

Minimum N is 6. Approximate value is calculated from the 6th partial sum.

Step-by-step solution

Understanding the Series

The series given is \( \frac{\pi}{\sqrt{12}} = \sum_{k=0}^{\infty} \frac{(-3)^{-k}}{2k+1} \). This is an alternating series and we need to find the minimum N such that the remainder is less than 0.0001.

Applying the Alternating Series Test

According to the alternating series test, the absolute value of the remainder \( R_N \) after \( N \) terms is less than or equal to the absolute value of the \( (N+1)^{th} \) term. Hence, for this series, \( |R_N| \leq \left| \frac{(-3)^{-(N+1)}}{2(N+1) + 1} \right| \).

Setting the Error Condition

We want \( \left| \frac{(-3)^{-(N+1)}}{2(N+1) + 1} \right| < 0.0001 \). This inequality will help us find the minimum value of \( N \) that ensures the error bound is respected.

Solving for N

We simplify the inequality: \( \frac{1}{3^{N+1}(2N+3)} < 0.0001 \). Rearranging gives us \( 3^{N+1}(2N+3) > 10000 \). We need to solve this inequality to find the smallest integer \( N \).

Trial and Error Method

Let's test a few values of \( N \):- For \( N = 5 \), \( 3^{6}(13) = 13 \times 729 = 9477 < 10000 \).- For \( N = 6 \), \( 3^{7}(15) = 15 \times 2187 = 32805 > 10000 \).Therefore, \( N = 6 \) is the smallest integer satisfying the condition.

Calculating the Partial Sum

Calculate the partial sum for \( N = 6 \):1. \( S_6 = \sum_{k=0}^{6} \frac{(-3)^{-k}}{2k+1} \).2. Compute each term and sum them to find \( S_6 \).

Finding the Approximate Value

The approximate value of \( \frac{\pi}{\sqrt{12}} \) is \( S_6 \) with an error less than 0.0001. Therefore, calculate the partial sum up to the 6th term to get this approximation.

Key concepts

Error Bound

Understanding the error bound is crucial in determining how much error we are willing to tolerate when estimating a sum. In the context of alternating series, the error bound relates to how closely the partial sum of the series approximates the infinite series.

In the given exercise, the error bound is specified as less than 0.0001. This means that the difference between the actual value of the infinite series and the partial sum that we calculate should not exceed this value. A clear understanding of the error bound ensures that the approximation is sufficiently close to the true value of the series. To calculate this, the \( N+1 \)-th term must have an absolute value smaller than the error bound, guiding us to determine the smallest possible \( N \) such that \(|R_N| < 0.0001\).

Partial Sum

A partial sum is the sum of the first few terms of a series. In this exercise, we are concerned with the partial sum of an alternating series.

Partial sum helps in approximating the total sum of an infinite series. For the series given in the exercise, the partial sum \( S_N \) is found by summing terms from the 0th to the \( N \)-th term. For example, to obtain \( S_6 \), you would add up all terms from \( k = 0 \) to \( k = 6 \) using the formula provided:

• \[ S_6 = \sum_{k=0}^{6} \frac{(-3)^{-k}}{2k+1} \]
• Compute each term and sum them to find \( S_6 \)

By calculating the partial sum, we not only approximate the value of the entire series but also ensure it meets the error tolerance defined by the error bound.

Alternating Series Test

The Alternating Series Test is a method used to determine the convergence of series where the signs of the terms alternate. In our problem, the series switches between positive and negative terms, making it an alternating series.

To apply the Alternating Series Test, two main conditions must be checked:

• The absolute value of the terms must be decreasing as you move through the series.
• The limit of the absolute value of the terms must approach zero as the number of terms tends toward infinity.

For the series \( \frac{\pi}{\sqrt{12}} = \sum_{k=0}^{\infty} \frac{(-3)^{-k}}{2k+1} \, \), these criteria are met.

Using the test, we determine that for \( N = 6 \,\) the remainder of the series (or the error) after six terms is adequately small, making it a reliable approximation to the actual sum of the infinite series.