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] \(b_{n}=1 / \sqrt{n}\), error \(<10^{-3}\)

Short answer

Use N = 999,999 to ensure the error is less than 0.001. Calculate \( S_{999,999} \) for the partial sum.

Step-by-step solution

Understanding the Alternating Series Estimation

For an alternating series of the form \( \sum_{n=1}^{\infty} (-1)^{n+1} b_n \), where \( b_n \) are positive, decreasing, and \( \lim_{n \to \infty} b_n = 0 \), the error \( \left| R_N \right| \) in using the first \( N \) terms to estimate the series satisfies \( \left| R_N \right| \leq b_{N+1} \).

Setting Up the Inequality

We need to find \( N \) such that \( b_{N+1} \leq 10^{-3} \). Given \( b_n = \frac{1}{\sqrt{n}} \), we set up the inequality \[ \frac{1}{\sqrt{N+1}} \leq 10^{-3}. \]

Solving the Inequality for N

Solve \( \frac{1}{\sqrt{N+1}} \leq 10^{-3} \) by first taking reciprocals, yielding \( \sqrt{N+1} \geq 10^3 \). Squaring both sides gives \( N+1 \geq 10^6 \), thus \( N \geq 10^6 - 1 = 999,999 \).

Finding the Partial Sum \( S_N \) for N = 999,999

Calculate the partial sum \( S_{999,999} = \sum_{n=1}^{999,999} (-1)^{n+1} \frac{1}{\sqrt{n}} \). Since this involves a large computation, it typically requires programming or computational software to sum this number of terms.

Key concepts

Error Estimation

Error estimation is key in analyzing the accuracy of an approximation when working with infinite series, particularly in alternating series.
The problem provides an upper bound for the error by using the estimate \( \left| R_N \right| \leq b_{N+1} \). This means that the true sum of the series will differ from the partial sum \( S_N \) by no more than \( b_{N+1} \).
To meet a given error tolerance, say \(< 10^{-3}\), we must select a suitable number of terms \( N \) such that \( b_{N+1} \) is small enough to ensure the error does not exceed this limit.

• For this specific problem with \( b_n = \frac{1}{\sqrt{n}} \), we ensure that the error satisfies \( \left| R_N \right| \leq \frac{1}{\sqrt{N+1}} \leq 10^{-3} \).
• This provides a way to determine a sufficiently large \( N \) so that our series approximation is precise within the desired error margin.

Partial Sum

A partial sum, denoted as \( S_N \), is the sum of the first \( N \) terms of a series.
In the context of an alternating series, the partial sum is used to approximate the infinite series by using a finite number of terms.
Calculating a partial sum helps measure how close our approximation is to the actual sum of the entire series. In this exercise, having computed that we need \( N = 999,999 \), the partial sum \( S_{999,999} \) can be expressed as:

• \( S_{999,999} = \sum_{n=1}^{999,999} (-1)^{n+1} \frac{1}{\sqrt{n}} \).
• Due to the sheer number of calculations, it's practical to use a computer program or software to compute \( S_{999,999} \) accurately.

Such computations ensure that athletes of partial sums are precise when working with very large series.

Infinite Series

An infinite series is a summation of an infinite sequence of terms.
Unlike a finite series, which has a definite number of terms, an infinite series consists of an endless number of terms being added.
This type of series often requires techniques to estimate or approximate the value of the infinite sum.

• Alternating series, like the one addressed in this exercise, alternate their signs in each term \((-1)^{n+1} b_n\), which can be beneficial when estimating and calculating partial sums.
• The behavior of infinite series, whether they converge to a finite value or diverge, is crucial to understanding and solving series problems.

Estimation and convergence are essential concepts when discussing infinite series, ensuring that any calculated value is not only reachable but also reasonably accurate.