Problem 48 - Essential Calculus

Question

Verify the sum. Then use a graphing utility to approximate the sum with an error of less than 0.0001. $$ \sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n !}\right)=\frac{e-1}{e} $$

Step-by-Step Solution

Answer

The sum of the series \( \sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n !}\right) \) converges to \( \frac{e-1}{e} \). This sum can be verified graphically with a margin of error less than 0.0001.

Step 1: Simplify the Infinite Sum

Note that \( (-1)^{n-1} \) alternates between positive and negative for each term. The factor \( \frac{1}{n!} \) continually gets smaller as \( n \) increases. So the general form of the infinite sum is as follows: \( \frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - \frac{1}{4!} + \dots \)

Step 2: Identify the Series Type

Recognize this as an alternating series where the series \( \frac{1}{n!} \) is the sequence of terms alternated in sign.

Step 3: Find the Sum of the Infinite Series

The sum of the series can be found from the formula for the sum of an infinite alternating series. Given that \( \frac{1}{n!} \rightarrow 0 \) as \( n \rightarrow \infty \), the alternating series test ensures the series converges to the sum \( S \) defined by \( S = a - \frac{a}{2} + \frac{a}{3} - \frac{a}{4} + \ldots \).

Step 4: Substitute Values and Simplify

In the case of our series, \( a = \frac{1}{1!} = 1 \). Hence, \( S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \), which simplifies to \( \frac{e-1}{e} \).

Step 5: Verify the Sum with a Graphing Utility

Utilize a graphing utility to plot the series and its sum. Adjust the parameters such that the error is less than 0.0001. Confirm that the graphical result matches the calculated result.

Key Concepts

Convergence of Series

When we talk about convergence in a series, we're looking to see if the sum of an infinite series approaches some finite value, or limit, as more terms are added. The series in question is the alternating series \[\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n !}\right)\]which is a sum of terms where the sign alternates between positive and negative. What guarantees that this series converges? This is where the Alternating Series Test comes into play.
To use the Alternating Series Test, two conditions must be met:

• The absolute value of the terms decreases monotonically: \( b_{n} \) is such that \( b_{n} = \left|\frac{1}{n!}\right| \), and for this series, \( b_{n+1} < b_{n} \) for all \( n \).
• The limit of the absolute value of the terms approaches zero: \( \lim_{n \to \infty} b_{n} = 0 \).

If both are satisfied, the series is guaranteed to converge. Factorials grow very quickly, meaning \( \frac{1}{n!} \) becomes incredibly small as \( n \) increases. Thus, \( b_{n} \) approaches zero. Therefore, the series converges, verifying the given sum.

Graphing Utility

A graphing utility can be an invaluable tool when working with series. It provides a visual representation of how a series behaves as more terms are added. When verifying the sum of the series \[\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n!}\right) = \frac{e-1}{e}\]a graphing utility helps to visually confirm the convergence and accuracy.To use a graphing utility in this context:

• Plot the partial sums of the series. These are sums of the first \( n \) terms of the series.
• As more terms are added, observe how the partial sums hold steady around the calculated sum \( \frac{e-1}{e} \).
• Continue increasing \( n \) until the difference between consecutive partial sums is less than 0.0001, the specified error tolerance.

By doing so, you confirm that as you add more terms, the partial sums get closer to the expected sum, supporting analytical computations with visual evidence.

Factorials

Factorials are an important mathematical concept denoted by \( n! \) and defined as the product of all positive integers up to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). The factorial of zero, \( 0! \), is defined as 1.
Factorials grow sharply as \( n \) increases, and this plays a crucial role in the convergence of series. Specifically, in the series\[\sum_{n=1}^{\infty}(-1)^{n-1}\left(\frac{1}{n!}\right) \]each term \( \frac{1}{n!} \) gets smaller as \( n \) increases due to the rapid growth of \( n! \).

• The factorial function makes terms in the series approach zero swiftly, satisfying one of the conditions for convergence.
• It dramatically reduces the size of each subsequent term, ensuring that the series converges even faster.

Understanding factorials helps us see why the series behaves as it does, as the factorial's rapid growth maintains convergence by keeping terms very small, leading to the finite sum \( \frac{e-1}{e} \).