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] \(b_{n}=1 / 2^{n}\), error \(<10^{-6}\)

Short answer

For \( N = 20 \), \( S_{20} \approx 0.666666 \).

Step-by-step solution

Understand the Alternating Series Estimation Theorem

The Alternating Series Estimation Theorem states that if a series \( \sum_{n=1}^{\infty} (-1)^{n+1} b_n \) satisfies the conditions for convergence, then the error \( |R_N| \) in approximating the sum by the first \( N \) terms, \( S_N = \sum_{n=1}^{N} (-1)^{n+1} b_n \), is at most \( |b_{N+1}| \). Our goal is to choose \( N \) such that \( |b_{N+1}| < 10^{-6} \).

Evaluate the Condition for Error

We need to find \( N \) so that \( b_{N+1} = \frac{1}{2^{N+1}} < 10^{-6} \). The inequality \( \frac{1}{2^{N+1}} < 10^{-6} \) gives a condition that \( 2^{N+1} > 10^{6} \).

Solve the Inequality for N

Calculate \( N \) by solving the inequality \( 2^{N+1} > 10^{6} \):1. Take the log of both sides, we get \((N+1) \log_{10}(2) > 6\).2. Calculate \( \log_{10}(2) \approx 0.3010 \), so we have \( N+1 > \frac{6}{0.3010} \approx 19.93 \).3. Thus, \( N+1 \approx 20 \), which implies \( N \geq 19 \). But since \( N+1 \) must be greater than 19.93, take \( N = 20 \).

Calculate the Partial Sum \( S_{N} \) for \( N = 20 \)

Now calculate the partial sum \( S_{20} \):\[ S_{20} = \sum_{n=1}^{20} (-1)^{n+1} \frac{1}{2^n} \].This is a geometric series with the first term \( a = \frac{1}{2} \) and a common ratio \( r = \frac{-1}{2} \). The partial sum can be computed using the formula for the sum of a geometric series: \[ S_{20} = \frac{1 - (-1/2)^{20}}{1 - (-1/2)} \]=\[ \frac{1 - (1/2)^{20}}{3/2} \].Evaluate this expression to approximate \( S_{20} \).

Simplify and Calculate \( S_{20} \)

Compute \( (1/2)^{20} \approx 9.5367 \times 10^{-7} \), then:\[ S_{20} = \frac{1 - 9.5367 \times 10^{-7}}{3/2} = \frac{1 - 9.5367 \times 10^{-7}}{1.5} \].Approximate \( 1 - 9.5367 \times 10^{-7} \approx 0.999999 \),So: \[ S_{20} \approx \frac{0.999999}{1.5} = 0.666666 \text{ or about } \frac{2}{3} \].

Key concepts

Geometric series

A geometric series is a type of infinite series where each term after the first is found by multiplying the previous term by a constant called the "common ratio." In our exercise, the sequence of terms is defined by the formula \( b_n = \frac{1}{2^n} \).
This is because each term is a power of \( \frac{1}{2} \), giving us a sequence like \( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \). The first term is \( a = \frac{1}{2} \) and the common ratio \( r \) is \( \frac{1}{2} \) in magnitude, which is the base of the exponent in \( b_n \).
The properties of geometric series are crucial because they allow us to derive formulas for sums, both finite and infinite, which are especially simpler when the series converges, as the absolute value of \( r \) is less than 1.

Error estimation

Error estimation is a critical part of series calculations, especially when dealing with the convergence of an infinite series. In the context of our problem, it refers to determining how close the partial sum \( S_N \) of the first \( N \) terms of a series is to the true sum of the infinite series.
Using the Alternating Series Estimation Theorem, we know that the error or the difference \( |R_N| \) is less than or equal to the absolute value of the first omitted term \( |b_{N+1}| \).
In practice, calculating this can help us choose an \( N \) such that \( |b_{N+1}| < 10^{-6} \), ensuring the desired precision or error bound is met when estimating the sum of the infinite series.

Infinite series

An infinite series is essentially the sum of the terms of an infinite sequence. Each term in the series builds upon the last, creating a potential to sum up infinitely.
Our exercise involves an infinite series denoted by \( \sum_{n=1}^{\infty}(-1)^{n+1}b_n \).
Infinite series come with rules and tests to determine their behavior, particularly whether they converge (sum approaches a finite value) or diverge (sum grows without bound).
Understanding whether an infinite series converges is fundamental to science and engineering, as it determines if the series can be effectively used in calculations and predictions.

Convergence of series

Convergence of a series tells us whether the sum of its infinite terms approaches a finite value. This characteristic is key to determining if the series can serve practical applications.
In the case of an alternating series like \( \sum (-1)^{n+1} b_n \), convergence is possible when the terms \( b_n \) decrease steadily to 0. This series in the exercise converges thanks to the geometric nature and the condition \( |r| < 1 \).
Checking for convergence involves understanding both the general form of the series and the nature of its terms. When a series converges, it can be approximated by a finite sum \( S_N \) with a known small error, which is highly useful in scientific computations.