Problem 13 - Essential Calculus Early Transcendentals

Question

\(13-16=\) Approximate the sum of the series correct to four decimal places. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2 n) !} $$

Step-by-Step Solution

Answer

Approximate series sum to four decimal places: -0.4597

Step 1: Understanding the Problem

The series presented is an alternating series of the form: \( \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n)!} \). Our task is to approximate the sum of this series to four decimal places.

Step 2: Identify the Alternating Series Test

The alternating series test states that if a series \( \sum_{n=1}^{\infty} (-1)^n a_n \) satisfies three conditions: 1) \( a_n > 0 \) for all \( n \), 2) \( a_n > a_{n+1} \) for all \( n \), and 3) \( \lim_{n \to \infty} a_n = 0 \), then the series converges. Here, \( a_n = \frac{1}{(2n)!} \), which meets all three conditions, suggesting convergence.

Step 3: Truncate the Series

Since the series is alternating and convergent, it can be approximated by truncating after a few terms. We calculate and then truncate at the point where the absolute value of the term is less than 0.0001, the precision required for four decimal places.

Step 4: Calculate Individual Terms

Calculate the first few terms: - For \( n=1 \), we have \( \frac{(-1)^{1}}{(2 \times 1)!} = -\frac{1}{2!} = -0.5 \).- For \( n=2 \), we have \( \frac{(-1)^{2}}{(2 \times 2)!} = \frac{1}{4!} = \frac{1}{24} \approx 0.04167 \).- For \( n=3 \), we have \( \frac{(-1)^{3}}{(2 \times 3)!} = -\frac{1}{6!} = -\frac{1}{720} \approx -0.00139 \).- For \( n=4 \), we have \( \frac{(-1)^{4}}{(2 \times 4)!} = \frac{1}{8!} = \frac{1}{40320} \approx 0.00002 \).

Step 5: Sum the Terms to Approximate

Sum these terms until changes are below the desired precision: - Partial sum for terms \( n=1 \) to \( n=3 \) is \( -0.5 + 0.04167 - 0.00139 \approx -0.45972 \). The fourth term \( 0.00002 \) affects the result less than the required precision of four decimal places, so it is ignored for this approximation.

Step 6: Verify Precision

The additional terms won't change the sum beyond four decimal places: Adding the term from \( n = 4 \) has an effect of \( +0.00002 \), which does not affect the first four decimal places of \( -0.4597 \). Thus, the series can be approximated to four decimal places as \( -0.4597 \).

Key Concepts

Convergence

Convergence is a fundamental concept in the study of series. It describes a situation where the terms of a series approach a specific value as the number of terms increases indefinitely. For a series to converge, it must have terms that diminish in value and eventually become negligible. This means as you add more terms, they have less and less effect on the overall sum.In our original exercise, we have an alternating series where the terms are given by \(a_n = \frac{1}{(2n)!}\). As \(n\) increases, \(a_n\) approaches zero because factorial in the denominator increases rapidly. This rapid increase ensures that each subsequent term contributes less to the sum, indicating convergence.

• Convergence means the series approaches a fixed value with increasing terms.
• The key condition for convergence is diminishing term size.
• Once a series converges, adding more terms won't significantly change its sum.

Recognizing convergence helps in calculating series sums accurately without having to consider infinitely many terms, as demonstrated in this problem.

Alternating Series Test

The Alternating Series Test (AST) is a specific type of test used to determine if an alternating series converges. An alternating series is one where the signs of the terms alternate between positive and negative, such as the series given by \( \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)!}\).For an alternating series \( \sum_{n=1}^{\infty} (-1)^n a_n \) to converge, it must satisfy three conditions:

• The terms \( a_n \) must be positive.
• Each term must be smaller than the preceding one, mathematically \( a_{n} > a_{n+1} \).
• The limit of the terms as \( n \to \infty \) should be zero, \( \lim_{n \to \infty} a_n = 0 \).

In the given exercise, \( a_n = \frac{1}{(2n)!} \) meets these conditions:

• Each \( a_n \) is positive for all \( n \).
• \( a_n > a_{n+1} \) because factorial functions increase rapidly in the denominator, making each term smaller than the previous term.
• \( \lim_{n \to \infty} a_n = 0 \), as explained by the nature of factorial growth.

Once an alternating series passes the AST, we know it converges and allows us to use approximations effectively.

Truncation of Series

Truncation of a series involves cutting off the series after a certain number of terms when calculating its sum. This process is especially useful when working with convergent alternating series because it allows us to approximate the infinite series by summing only a finite number of terms. In essence, you stop adding terms to the series once subsequent terms become so small that they do not affect the result up to the required precision. In our example, this is four decimal places.

• Truncate when additional terms add less than the desired precision (e.g., 0.0001 for four decimal places).
• Calculate just enough terms to achieve the needed precision.
• Ignore terms beyond the cutoff point since they contribute insignificantly.

By truncating our series as demonstrated in the solution, we calculated an accurate approximation while maintaining efficiency in our calculations. This technique simplifies finding series sums to any desired level of precision without exhaustive computation.