Problem 57 - Thomas Calculus

Question

How many terms of the convergent series \(\sum_{n=1}^{\infty}\left(1 / n^{1.1}\right)\) should be used to estimate its value with error at most 0.00001\(?\)

Step-by-Step Solution

Answer

Around 14,125,315 terms are needed.

Step 1: Understanding the Problem

You are asked to find out how many terms of the series \(\sum_{n=1}^{\infty}\left(1 / n^{1.1}\right)\) need to be summed to achieve an error less than 0.00001. The series is convergent.

Step 2: Comparing with p-series

Recognize that \(\sum_{n=1}^{\infty}\left(1 / n^{p}\right)\) is a convergent series when \(p>1\). Here, \(p=1.1\), which means our series is convergent.

Step 3: Applying the Integral Test

The Integral Test helps estimate the error. For a convergent \(p\)-series, the remainder \(R_n = \int_{n}^{\infty}\frac{1}{x^{1.1}}dx\) gives the error.

Step 4: Calculating the Integral

Evaluate \(\int_{n}^{\infty}\frac{1}{x^{1.1}}dx = \left.\frac{x^{-0.1}}{-0.1}\right|_{n}^{\infty} = \frac{10}{9}\cdot n^{-0.1}\).

Step 5: Setting Error Inequality

Set \(\frac{10}{9}n^{-0.1} < 0.00001\) and solve for \(n\).

Step 6: Solving the Inequality

Rearrange to get \(n^{-0.1} < \frac{10^{-4}}{\frac{10}{9}}\) which simplifies to \(n^{0.1} > \frac{9}{10000}\). Calculating this gives \(n > \left(\frac{9}{10000}\right)^{-10}\).

Step 7: Final Calculation

Evaluate \(n > \left(\frac{9}{10000}\right)^{-10}\) to get the number of terms \(n\). In this case, \(n \approx 14125315\).

Key Concepts

Integral Test

The Integral Test is a fundamental tool in calculus used to assess the convergence of infinite series. It connects the worlds of calculus and series, allowing us to examine whether a series converges by comparing it to an improper integral.
The key idea of the Integral Test is:

• If the function associated with the series is positive, continuous, and decreasing, then the series and the corresponding improper integral either both converge or both diverge.

In practice, for a series \sum_{n=1}^{\infty} a_n\, we consider the integral of the function f(x), where a_n = f(n). We then evaluate the integral \int_{n}^{\infty} f(x)\, to determine the behavior of the series.
In the exercise, the integral test is applied to the series \(\sum_{n=1}^{\infty}\left(\frac{1}{n^{1.1}}\right)\) to calculate the error in estimating the sum of the series by finite terms. By substituting the infinite series with an improper integral, it becomes much easier to handle and find an error estimate.

P-Series

P-series are mathematical series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\).
These series are pivotal in understanding series convergence, as they provide a benchmark for comparison with other series.

• When \(p > 1\), a p-series converges
• When \(p \leq 1\), a p-series diverges.

The series in the exercise \(\sum_{n=1}^{\infty}\left(\frac{1}{n^{1.1}}\right)\) is an example of a convergent p-series because \(p = 1.1\) is greater than 1.
Understanding p-series helps predict the behavior of more complex series by categorizing them based on their defining exponent \(p\). If a series can be manipulated or approximated into a form of p-series, its convergence properties can often be determined.

Series Convergence

Series convergence refers to the condition when the sum of an infinite series approaches a finite value as the number of terms increases. Convergence is a crucial concept in calculus, as it defines whether a series can be used for precise calculations.

• To determine if a series converges, various tests can be applied including the integral test, comparison test, and ratio test, among others.
• Convergence is dependent on both the general term \(a_n\) and the behavior of the series as \(n\) becomes large.

For the series \(\sum_{n=1}^{\infty}\left(\frac{1}{n^{1.1}}\right)\) in the exercise, convergence is guaranteed by the p-series criteria: \(p = 1.1\) indicates that the series converges, aligning with the integral test result.
This guarantees that given a large enough number of terms, the series will approximate a specific value, allowing it to be utilized in practical calculations.

Error Estimation

Error estimation is crucial when summing series to a finite number of terms, as it offers a measure of how close the partial sum of a series is to its actual total sum. In calculus, understanding the possible discrepancy between the partial sum \(S_n\) and the total sum \(S\) is important for precise calculations.
The exercise implements error estimation using the formula derived from the integral test. Specifically for a p-series, the remainder \(R_n\) (which is analogous to the error) is calculated using the improper integral\(R_n \leq \int_{n}^{\infty} \frac{1}{n^{1.1}}\). Using this, a boundary is established on the error made by summing \(n\) terms instead of infinitely many.

• The integral approximates \(R_n\), enabling the determination of how many terms \(n\) are needed to keep this error less than a desired threshold, like 0.00001 in the exercise.

This accurate method of estimation ensures that the series value can be approached within any given margin of error by summing a sufficient number of terms.