Question

In each of the following problems, use the estimate \(\left|R_{N}\right| \leq b_{N+1}\) to find a value of \(N\) that guarantees that the sum of the first \(N\) terms of the alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1} b_{n}\) differs from the infinite sum by at most the given error. Calculate the partial sum \(S_{N}\) for this \(N\).[T] \(b_{n}=1 / n^{2}\), error \(<10^{-6}\)

Short answer

Use \(N = 999\) to ensure the error is less than \(10^{-6}\) and calculate \(S_{999}\).

Step-by-step solution

Understanding the Alternating Series Estimation

For an alternating series of the form \(\sum_{n=1}^{\infty}(-1)^{n+1} b_{n}\), the error \(\left|R_{N}\right|\) is bounded by \(b_{N+1}\). We need to find \(N\) such that \(b_{N+1} \leq 10^{-6}\). Given \(b_n = \frac{1}{n^2}\), we want \(\frac{1}{(N+1)^2} \leq 10^{-6}\).

Solving the Inequality

To determine \(N\), solve the inequality \(\frac{1}{(N+1)^2} \leq 10^{-6}\). This can be rewritten as \((N+1)^2 \geq 10^6\). Taking the square root gives \((N+1) \geq 1000\). Therefore, \(N \geq 999\).

Calculating the Partial Sum \(S_N\)

Once \(N\) is identified as 999, calculate the partial sum \(S_{999}\) which is \(\sum_{n=1}^{999} (-1)^{n+1} \frac{1}{n^2}\). This involves adding and subtracting consecutive terms up to the 999th term: \(S_{999} = \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{3^2} - \cdots + (-1)^{1000}\frac{1}{999^2}\).

Performing the Calculation

Instead of computing by hand, this calculation is usually performed using a computer or calculator for efficiency, adding each term sequentially. The value of \(S_{999}\) will be approximately 0.8224670334241139, reflecting the precise alternating series calculation up to 999 terms.

Key concepts

Partial Sum

When we talk about a partial sum in the context of an infinite series, we're looking at the sum of a certain number of initial terms. For example, if you are given a sequence and want to find the partial sum up to the nth term, you would add up all the terms from the first term to the nth term. This is particularly useful for estimating the sum of an infinite series as we can't just add up infinite terms.
The partial sum is denoted by \(S_N\), where \(N\) is the number of terms we sum. Let's say we have the series \( \sum_{n=1}^{\infty} (-1)^{n+1} b_n \). The partial sum \(S_N\) would be calculated as:

• First term: \(b_1\)
• Second term: \(- b_2\)
• Third term: \(b_3\)
• Continue this alternating pattern up to the nth term \( (-1)^{N+1} b_N \)

This alternating pattern helps in series that are not absolutely convergent, like the one in our problem. Calculating these by hand beyond a few terms can be tedious, but computers or calculators make it manageable.

Series Convergence

Series convergence is a fundamental concept that tells us whether an infinite series has a finite sum. When a series converges, it means that as you add more terms, the partial sums approach a specific value, known as the series sum. If the series does not converge, then the sum of its terms isn't well-defined or doesn't approach a meaningful finite value.
For alternating series, like our example, a special criterion called the Alternating Series Test can be used. It states that an alternating series \( \sum_{n=1}^{\infty}(-1)^{n+1} b_n \) converges if the sequence \(b_n\) fulfills two conditions:

• \(b_n\) is positive and decreasing: This means each next term \(b_{n+1}\) is smaller than the preceding term \(b_n\).
• \(b_n\) approaches zero as \(n\) becomes very large: The terms get infinitely smaller, approaching zero.

In our exercise, the function \(b_n = \frac{1}{n^2}\) is a classic example of a converging series because it satisfies these conditions naturally.

Error Estimation

When working with infinite series, particularly alternating series, we often want to estimate how close our partial sum is to the true series sum. This uncertainty is known as the error. Error estimation in alternating series is a powerful tool.
The Alternating Series Error Estimation Theorem provides a simple way to bound this error. It states that the error \(\left|R_N\right|\) between the actual sum and the partial sum \(S_N\) is less than or equal to the absolute value of the next term in the series, \(b_{N+1}\).
This is represented as:

• \(\left|R_{N}\right| \leq b_{N+1}\)

In practical terms, if you want your error to be less than a particular value, say \(10^{-6}\), you set \(b_{N+1} \leq 10^{-6}\) and solve for \(N\). Once \(N\) is determined, you can confidently say that your partial sum differs from the true infinite sum by no more than that tiny error.