Problem 52 - Thomas Calculus

Question

How many terms of the convergent series \(\sum_{n=4}^{\infty}\left(1 / n(\ln n)^{3}\right)\) should be used to estimate its value with error at most 0.01\(?\)

Step-by-Step Solution

Answer

Use 1174 terms to estimate the series with error at most 0.01.

Step 1: Identify the Error Term Formula

The problem asks for the number of terms required to estimate the sum of the series with an error less than or equal to 0.01. For a convergent series, we often use the integral test to find an upper bound of the remainder term. The error is given by the remainder of the series, and for a series like \(\sum_{n=4}^{\infty} a_n\), we use the integral \(\int_{n}^{\infty} a(x)\,dx\) as an approximation for the error of the series sum.

Step 2: Set Up the Integral Test

To apply the integral test, consider the function \(f(x) = \frac{1}{x(\ln x)^3}\) which is a continuous, positive, and decreasing function for \(x > 3\). We set up the integral \(\int_{n}^{\infty} \frac{1}{x (\ln x)^3}\,dx\) to approximate the remainder (error).

Step 3: Solve the Integral

Calculate the integral \(\int_{n}^{\infty} \frac{1}{x (\ln x)^3}\,dx\). To solve, perform a substitution: let \(u = \ln x\) then \(du = \frac{1}{x} dx\). The integral becomes \(\int \frac{1}{u^3}\, du = -\frac{1}{2u^2} + C\). Thus, \[\int_{n}^{\infty} \frac{1}{x (\ln x)^3}\,dx = -\frac{1}{2 (\ln n)^2}. \]

Step 4: Set the Remainder Less Than 0.01

We are given that the error should be at most 0.01. Therefore, we set the inequality \(-\frac{1}{2(\ln n)^2} < 0.01\). Solving this inequality will give us the value of \(n\) such that the error is less than 0.01.

Step 5: Solve the Inequality

Simplify and solve the inequality \(-\frac{1}{2 (\ln n)^2} < 0.01\), which becomes \((\ln n)^2 > 50\). Solving further, we get \(\ln n > \sqrt{50}\) and then taking exponential on both sides gives us \(n > e^{\sqrt{50}}\).

Step 6: Calculate the Value of \(n\)

Calculate \(n > e^{\sqrt{50}}\). Approximating \(\sqrt{50}\) gives approximately 7.07, thus \(n > e^{7.07} \approx 1173.43\). Hence, the nearest integer value of \(n\) that satisfies the inequality is 1174.

Key Concepts

Integral Test

The Integral Test is an essential tool in determining the convergence of an infinite series. If you have a series \( \sum_{n=1}^{\infty} a_n \), and you want to determine its behavior, the Integral Test states that you can find a function \( f(x) \) that matches \( a_n \).
This function should satisfy three conditions: it must be positive, continuous, and decreasing on the interval from \( n \) to \( \infty \).

• Positive: The function must yield positive values for all \( x > c \).
• Continuous: There must be no breaks in the function over the interval.
• Decreasing: The function must be steadily decreasing as \( x \) increases.

In practice, this means checking whether \( f(x) \) decreases as more terms are added.
If the integral of \( f(x) \) from \( c+1 \) to \( \infty \) converges, then the series \( \sum_{n=1}^{\infty} a_n \) converges as well. In the original exercise, using the integral test helped confirm that the series converged, aiding in determining the required number of terms for a specific error.

Error Estimation

When dealing with convergent series, estimating the error can give you an upper bound for the approximation. This estimation plays a crucial role in knowing how many terms you need to include to achieve accurate results. The integral test not only tells us about convergence but is also used for estimating errors.
In our case, after applying the integral, we find an expression that describes the potential error when estimating the sum with a specified number of terms. This involves calculating the integral from \( n \) to \( \infty \) instead of considering countless terms, which is more practical.

• The integral gives an estimate for the sum beyond the term \( n \).
• Knowing and controlling the error allows accurate approximation of series up to desired levels, such as 0.01 in the exercise.

Essentially, error estimation simplifies the task significantly by providing a manageable way to understand how our approximation compares to the actual sum value.

Series Remainder

The remainder of a series refers to the total of all remaining terms after a certain number of terms have been summed. Knowing how to handle remainders is vital when summing infinite series, as it directly correlates to error estimation and affects the accuracy of approximations.
In the context of the given exercise, after identifying the series remainder using integration, you can control the error margin. The integral gives a framework to estimate how the remainder behaves and provides assurances on how many terms are necessary to satisfy an inequality for error control.

• The remainder helps determine the necessary number of terms to achieve a specified accuracy.
• By bounding the remainder, you ensure that your partial sum does not exceed a given error threshold, such as 0.01.

Thus, understanding series remainder through integral estimation becomes critical in approximating sums without summing all terms officially.