Problem 39 - Calculus: Early Transcendentals

Question

Estimate \( \sum_{n = 1}^{\infty} (2n + 1)^{-6} \) correct to five decimal places.

Step-by-Step Solution

Answer

Use integral approximation and partial sums to estimate the series.

Step 1: Understand the Series

The series we need to estimate is \( \sum_{n=1}^{\infty} (2n + 1)^{-6} \). It consists of terms given by \((2n+1)^{-6}\), which are a function of natural numbers \(n\). This is a type of series called an infinite series.

Step 2: Determine Convergence

We need to determine if the series converges or diverges. Since each term is positive and decreases, one can check the convergence using the p-series test. An infinite series \(\sum_{n=1}^{\infty} a_n\) where \(a_n = (2n+1)^{-p}\) converges if \(p>1\). Here, \(p=6\), which means the series converges.

Step 3: Use Integral Approximation

To approximate the sum of this series to five decimal places, we can use integral approximation. The integral of the function \(f(x) = (2x+1)^{-6}\) can be used to approximate the tail of the series. The integral from \(n\) to \(\infty\) of \(f(x)\) gives us an estimate of \(R_n\), the remainder after \(n\) terms.

Step 4: Choose an Integral

The integral that approximates \(\int_n^\infty (2x+1)^{-6} \, dx\) gives us the remainder \(R_n\). Solving this integral will allow us to calculate the tail of the series. Solving this requires knowledge of integration by substitution.

Step 5: Compute Numerical Terms

Compute the sum of the series for a reasonable number of finite terms to get the partial sum. Let's say we calculate up to \( n = N \), where the choice of \(N\) becomes critical to ensure that the calculated remainder gives an accurate total sum.

Step 6: Solve the Integral

Substituting \(u = 2x + 1\), we get the integral \(\int \frac{1}{u^6} \, du\) which equals \(-\frac{1}{5u^5} + C\). Evaluating this integral from \(N\) to \(\infty\) gives \(-\frac{1}{5(2N+1)^5}\) as a correction factor to add to the partial sum.

Step 7: Calculate Total Sum

Calculate the partial sum \(S_N = \sum_{n=1}^{N} (2n+1)^{-6}\) and add the remainder term \(R_N = \frac{1}{5(2N+1)^5}\) for the total estimate of the infinite series. Adjust \(N\) to ensure precision to five decimal places.

Key Concepts

Convergence Criteria

When analyzing an infinite series, determining if it converges or diverges is a crucial first step.
A series converges if the sum approaches a specific finite number. Conversely, it diverges if the sum grows indefinitely.
For the series \( \sum_{n=1}^{\infty} (2n+1)^{-6} \), we know each term diminishes as \( n \) increases. This is because the expression \((2n + 1)^{-6}\) decreases with larger values of \( n \).
One common test used is the p-series test, which states: an infinite series of the form \( \sum_{n=1}^{\infty} a_n \), where \( a_n = \frac{1}{n^p} \), converges if \( p > 1 \).
In our given series, substituting \( (2n+1)^{-p} \), with \( p = 6 \), we find that it satisfies this p-series criterion, confirming convergence.

P-Series Test

The p-series test is a method used to determine the convergence of series that involve terms raised to a power.
Specifically, a series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \) can be quickly assessed for convergence by looking at the exponent \( p \).

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

Our task is to analyze the series \( \sum_{n=1}^{\infty} (2n+1)^{-6} \). Applying the p-series test means we consider the general term \( (2n+1)^{-p} \) where \( p = 6 \).
Since \( 6 > 1 \), clearly the p-series test helps us confirm that this series indeed converges.
It's a very efficient and powerful tool for quickly determining the behavior of infinite series with similar structures.

Integral Approximation

Integral approximation comes into play when estimating the sum of a convergent series, especially when the exact sum is hard to find directly.
The integral test can aid in evaluating how close a partial sum of a series is to its infinite sum by estimating the tail, or the remaining series not included in the partial sum.
For a function \( f(x) \) that is continuous, positive, and decreasing on \([n, \infty)\), the remainder \( R_n \) can be approximated by the integral\[ R_n \approx \int_n^\infty f(x) \, dx\] For our specific series \( \sum_{n=1}^{\infty} (2n + 1)^{-6} \), we use \( f(x) = (2x+1)^{-6} \). By integrating \( f(x) \) from \( n \) to infinity, we get an estimate for this remainder.
Performing integration by substitution, we let \( u = 2x + 1 \). This simplification helps us find \( R_n \). The integral becomes relatively straightforward, aiding in the precision of our series estimate.

Remainder Estimation

In series calculations, remainder estimation allows us to assess the accuracy of a partial sum’s approximation of the total sum.
Once you have calculated the sum for a finite number of terms in a convergent series, estimating the remainder becomes essential to getting an accurate total.

• First, compute a partial sum \( S_N = \sum_{n=1}^{N} (2n+1)^{-6} \).
• Then, estimate the remainder \( R_N \), the series tail from \( N+1 \) to infinity.

Using integral approximation, the remainder \( R_N \) is estimated by the integral \( \int_N^\infty (2x+1)^{-6} \, dx \).
Substituting here, we solve: \[ R_N = \frac{1}{5(2N+1)^5}\] Adding \( R_N \) to \( S_N \) provides the complete estimated sum. Adjust \( N \) until this sum stabilizes to the required decimal precision, ensuring the sum is correct to five decimal places.