Problem 55 - Thomas Calculus

Question

Estimate the value of \(\sum_{n=1}^{\infty}\left(1 / n^{3}\right)\) to within 0.01 of its exact value.

Step-by-Step Solution

Answer

The series sums to approximately 1.1975, accurate to within 0.01.

Step 1: Understand the Series

The given series is called the Riemann zeta function, defined as \( \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \), for \( s > 1 \). Here, \( s = 3 \). The series converges because \( 3 > 1 \). We want to approximate \( \zeta(3) \) to within 0.01.

Step 2: Use Approximate Series

Since we want an approximation, compute partial sums of the series: \( S_k = \sum_{n=1}^{k} \frac{1}{n^3} \). Increase \( k \) until the remainder of the series, estimated by an appropriate method, is less than 0.01.

Step 3: Choose Remainder Estimation Method

We use the integral test for convergence to estimate the remainder: \( R_k = \int_{k+1}^{\infty} \frac{1}{x^3} \, dx \). Calculate this to find the error in using \( S_k \) as an approximation to \( \zeta(3) \).

Step 4: Compute the Integral for Remainder

Calculate the integral: \( \int_{k+1}^{\infty} \frac{1}{x^3} \, dx = \left[-\frac{1}{2x^2}\right]_{k+1}^{\infty} = \frac{1}{2(k+1)^2} \). Set this less than 0.01 to solve for \( k \).

Step 5: Solve Inequality for k

The inequality is \( \frac{1}{2(k+1)^2} < 0.01 \). Solving gives \( k+1 > \sqrt{50} \approx 7.07 \). Thus, \( k \geq 7 \).

Step 6: Calculate Partial Sum S_7

Compute the partial sum \( S_7 = \sum_{n=1}^{7} \frac{1}{n^3} \). This sum adds up to \( 1 + \frac{1}{8} + \frac{1}{27} + \frac{1}{64} + \frac{1}{125} + \frac{1}{216} + \frac{1}{343} \approx 1.1975 \).

Step 7: Verify Accuracy

Calculate the actual error: since \( R_7 = \frac{1}{2(8)^2} = \frac{1}{128} \approx 0.0078 \), the partial sum \( S_7 \) is an accurate estimate with error less than 0.01.

Key Concepts

Series Convergence

The concept of series convergence is fundamental when dealing with infinite series like the Riemann zeta function. When we say a series converges, it means its terms approach a finite number as more and more terms are added. For the Riemann zeta function, defined as \( \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} \), convergence happens when \( s > 1 \). In our original exercise, since \( s = 3 \), the series converges.
This is crucial as series that don't converge will not sum up to a finite value. Instead, they keep increasing indefinitely. For students, it's important to understand that the convergence of a series like the one with \( s = 3 \) allows us to calculate an approximate sum and makes it possible to estimate how close our approximation is to this actual value.

Partial Sum

A partial sum represents the sum of a finite number of terms from an otherwise infinite series. In our exercise, we use partial sums to approximate the total value of the Riemann zeta function for \( s = 3 \).
The partial sum is denoted by \( S_k = \sum_{n=1}^{k} \frac{1}{n^3} \), where \( k \) is the number of terms we choose to add up. Increasing \( k \) generally yields a more accurate approximation of \( \zeta(3) \).

• A smaller \( k \): Rough approximation.
• A larger \( k \): Better approximation.

By calculating these sums, students can see how the partial sum approaches the actual value as more terms are included. It gives a practical way to understand convergence and how closely it mimics the behaviour of the full series.

Remainder Estimation

After computing a partial sum, we can still be off from the true value of an infinite series. This "leftover" part is quantified as the remainder. The remainder estimation tells us how much we miss by not including all terms of the series beyond a certain point.
In the given exercise, we need to estimate the value of the series to within an error of 0.01. To manage this, we use a method called remainder estimation where we compute to see if \( R_k \), the remainder, is small enough. The key is ensuring this remainder, the integral from \( k+1 \) to infinity, is less than 0.01. Once it is, our partial sum is a good approximation of the exact sum.

Integral Test

The integral test is a convergence test that compares a series to an improper integral to determine the series' behaviour. Here, it also assists in estimating the remainder. If we express this series as a function \( f(x) = \frac{1}{x^3} \), the integral test helps us conclude that the series converges.
By calculating the integral \( \int_{k+1}^{\infty} \frac{1}{x^3} \, dx \), we gain insights into how quickly or slowly the terms of the series diminish after some point \( k \). This integral gives a formula for estimating the remainder \( R_k = \frac{1}{2(k+1)^2} \).
When applying it, you get an idea of how accurately the partial sum \( S_k \) represents the exact value of\( \zeta(3) \). Such estimation is handy because it provides a mathematical guarantee on the error's size and ensures that the approximation meets defined accuracy needs.