Question

Find the minimum value of \(N\) such that the remainder estimate \(\int_{N+1}^{\infty} f

Short answer

Practically impossible; using computational tools or approximation is necessary for an efficient solution.

Step-by-step solution

Understand the Error Estimate

We need to find an integer \( N \) such that the remainder term \( R_N \) satisfies the inequality \( R_N < 10^{-3} \). The error estimate for an infinite series \( \sum_{n=1}^{\infty} a_n \) is given by comparing it with the integral \( \int_{N}^{\infty} f(x) \, dx \), where \( f(x) = \frac{1}{x \ln^2 x} \) in this case.

Set Up the Integral

Given \( f(x) = \frac{1}{x \ln^2 x} \), we need to set up the integral \( \int_{N}^{\infty} \frac{1}{x \ln^2 x} \, dx \) and ensure that it is less than the given error of \( 10^{-3} \).

Evaluate the Integral

The integral \( \int \frac{1}{x \ln^2 x} \, dx \) can be evaluated using substitution. Let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \). This transforms the integral to \( \int \frac{1}{u^2} \, du = -\frac{1}{u} + C \). Evaluating from \( N \) to \( \infty \), we have \( \int_{N}^{\infty} \frac{1}{x \ln^2 x} \, dx = \left[ -\frac{1}{\ln x} \right]_{N}^{\infty} = \frac{1}{\ln N} \).

Solve the Inequality

Set \( \frac{1}{\ln N} < 10^{-3} \). This implies \( \ln N > 1000 \). Exponentiating both sides gives \( N > e^{1000} \), but since this is impractical, we must reconsider our approach by evaluating the value of \( N \) numerically for more reasonable errors or approximation techniques, as maximizing this function explicitly is practically impossible without computational tools.

Key concepts

Error Estimate

When working with infinite series, the error estimate helps us understand how accurate an approximation is when we cut off our summation at a certain point. Specifically, for the series \( \sum_{n=1}^{\infty} a_n \), where each term \( a_n = \frac{1}{n \ln^2 n} \), we achieve a better approximation by summing up to a certain number \( N \) and estimating the error that this truncation introduces.

The error estimate is akin to checking how much of the series remains beyond our chosen point \( N \). In this case, we aim to have this remaining error, or the remainder term \( R_N \), less than \( 10^{-3} \). This ensures the series is accurate enough for practical purposes.

To determine \( N \), we compare the remainder, which is what is left of the series after \( N \), with an integral. This concept is fundamental because it provides a systematic method for controlling the error in approximations.

Integral Comparison

Integral comparison is a valuable technique for estimating errors in infinite series. By comparing a series to an improper integral, we can gauge the size of the remaining terms when we truncate the series.

For \( f(x) = \frac{1}{x \ln^2 x} \), we consider the integral \( \int_{N}^{\infty} \frac{1}{x \ln^2 x} \, dx \). This integral represents our estimation of the terms left in the series from \( N+1 \) to infinity. Practically, when integrating \( f(x) \), we find that:

• Using substitution such as \( u = \ln x \), and \( du = \frac{1}{x} \, dx \), simplifies the integral to \( \int \frac{1}{u^2} \, du \).
• Upon evaluation, this integral results in the expression \( -\frac{1}{\ln x} \), spanning from \( N \) to infinity.

This result helps us conclude the comparison by determining that if this integral is less than one-thousandth (\( 10^{-3} \)), the terms we've neglected provide only a negligible margin of error in our series approximation.

Remainder Term Analysis

Analyzing the remainder term is all about understanding and controlling the leftover part of the infinite series that wasn't included in the finite sum. Here's how you visualize it: once you reach the term \( N \) in your series sum, you calculate what's left over if you were to continue summing infinite terms.

In remainder term analysis, after performing the integral and finding the expression \( \frac{1}{\ln N} \), you'll need this value to adhere to the condition \( \frac{1}{\ln N} < 10^{-3} \). This guides how we choose \( N \).

Solving the Inequality

You might find the solution \( \ln N > 1000 \) daunting. It implies \( N > e^{1000} \), a value far from practical for computation. In reality, estimating \( e^{1000} \) exactly is often unfeasible without computational help, so considerations for error might need adjustments depending on practical application or additional computational resources.

• Understanding that sometimes direct calculations aren't pragmatically achievable helps in seeking alternatives.
• Numerical methods or approximation algorithms may offer a path forward.

Thus, the remaining challenge is choosing \( N \) wisely when exact solutions become impractical, especially with immense numbers like \( e^{1000} \). Therefore, remainder term analysis becomes vital in ensuring a balance between error tolerance and computational feasibility.