Problem 33 - Calculus: Early Transcendentals

Question

Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error. $$ \displaystyle\sum_{n = 1}^{\infty} \frac {1}{5 + n^5} $$

Step-by-Step Solution

Answer

The sum of the first 10 terms estimates the series sum as ~0.1691, with an error less than 0.000006.

Step 1: Identify the Series

The series in question is \( \sum_{n=1}^{\infty} \frac{1}{5 + n^5} \). This is an infinite series, and we are tasked with approximating its sum using the first 10 terms.

Step 2: Calculate the First 10 Terms

To find the sum of the first 10 terms, compute each term \( a_n = \frac{1}{5 + n^5} \) for \( n = 1 \) to \( n = 10 \), then add them together.

Step 3: Determine Each Term

Calculate:- For \( n=1 \), \( a_1 = \frac{1}{6} \).- For \( n=2 \), \( a_2 = \frac{1}{37} \).- For \( n=3 \), \( a_3 = \frac{1}{248} \).- Continue this process up to \( n=10 \).

Step 4: Compute the Partial Sum

Add up all the terms from \( n=1 \) to \( n=10 \): \( S_{10} = \frac{1}{6} + \frac{1}{37} + \frac{1}{248} + \cdots + \frac{1}{10000005} \). Calculate this sum to get the approximate sum of the first 10 terms.

Step 5: Estimate the Error

For series where terms decrease and are positive, the error in approximating the sum by the first \( n \) terms is less than the first omitted term, \( a_{11} = \frac{1}{5 + 11^5} \).Calculate \( a_{11} \) to estimate the error.

Step 6: Calculate and Conclude

Determine the numerical value of \( S_{10} \) and \( a_{11} \) to get your estimates:- Use a calculator to find \( S_{10} \).- Calculate \( a_{11} = \frac{1}{11^5 + 5} = \frac{1}{161051 + 5} = \frac{1}{161056} \). Now you have the sum estimate and the estimated error.

Key Concepts

Approximation of Series

Approximating an infinite series means finding a finite sum that is close to the actual sum of the series. In many cases, we can't find the exact sum of a series that goes to infinity. Instead, we use part of the series to get as close as possible.

For the series in the problem, \( \sum_{n=1}^{\infty} \frac{1}{5 + n^5} \), we are interested in approximating it using only the first 10 terms. Each term of the series can be calculated as \( a_n = \frac{1}{5 + n^5} \), and these terms will get smaller as \( n \) becomes larger.

• This type of series focuses on a sequence that quickly shrinks as \( n \) increases.
• Since the terms decrease, the sum gets closer to the actual total sum of the infinite series with more terms.

Breaking an infinite series into a partial sum as we did here is a common technique, especially for complex series where finding an exact value is impractical.

Partial Sums

A partial sum is the sum of a finite number of terms of a series. Here, for approximating the infinite series, we computed the sum of the first 10 terms.

The partial sum \( S_{10} \) is given by adding up the first 10 terms, calculated individually as follows:

• \( a_1 = \frac{1}{6} \)
• \( a_2 = \frac{1}{37} \)
• \( a_3 = \frac{1}{248} \)
• ...continue this up to \( a_{10} = \frac{1}{10000005} \)

Summing these terms results in the partial sum \( S_{10} \), which provides us an approximation of the entire series. The more terms we include from the series, the more accurate our approximation will be, since partial sums converge to the actual sum as more terms are added.

Error Estimation

When approximating the sum of an infinite series with a partial sum, it's important to estimate how much error there is in our approximation. Essentially, we want to know how far off we might be from the real sum.

For series where the terms are positive and decrease, like the one in question, the error in approximating by the first \( n \) terms is less than the next term we did not use. Therefore, the error can be estimated using the first omitted term, which for \( n=10 \) is \( a_{11} = \frac{1}{5 + 11^5} \).

• This error calculation helps us understand the precision of our approximation.
• In mathematical terms, the remainder \( R_n \) is less than or equal to \( a_{11} \), which allows us to conclude that the sum of the series is within \( S_{10} \pm a_{11} \).

With this method, you get not only an approximation but also confidence about how close you are to the original infinite series' sum.