Problem 6 - Essential Calculus Early Transcendentals

Question

Calculate the first eight terms of the sequence of partial sums correct to four decimal places. Does it appear that the series is convergent or divergent? $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n !}$$

Step-by-Step Solution

Answer

The series appears to be convergent.

Step 1: Understanding the Series

The series given is \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!}\). We need to calculate the partial sums \(S_n = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k!}\) for \(n = 1, 2, \ldots, 8\).

Step 2: Calculate the First Term

For \(n = 1\), the first term is \(\frac{(-1)^{1-1}}{1!} = \frac{1}{1} = 1\). Thus, \(S_1 = 1\).

Step 3: Calculate the Second Term

For \(n = 2\), the second term is \(\frac{(-1)^{2-1}}{2!} = \frac{-1}{2} = -0.5\). Thus, \(S_2 = 1 - 0.5 = 0.5\).

Step 4: Calculate the Third Term

For \(n = 3\), the third term is \(\frac{(-1)^{3-1}}{3!} = \frac{1}{6} \approx 0.1667\). Thus, \(S_3 = 0.5 + 0.1667 = 0.6667\).

Step 5: Calculate the Fourth Term

For \(n = 4\), the fourth term is \(\frac{(-1)^{4-1}}{4!} = \frac{-1}{24} = -0.0417\). Thus, \(S_4 = 0.6667 - 0.0417 = 0.6250\).

Step 6: Calculate the Fifth Term

For \(n = 5\), the fifth term is \(\frac{(-1)^{5-1}}{5!} = \frac{1}{120} \approx 0.0083\). Thus, \(S_5 = 0.6250 + 0.0083 = 0.6333\).

Step 7: Calculate the Sixth Term

For \(n = 6\), the sixth term is \(\frac{(-1)^{6-1}}{6!} = \frac{-1}{720} \approx -0.0014\). Thus, \(S_6 = 0.6333 - 0.0014 = 0.6319\).

Step 8: Calculate the Seventh Term

For \(n = 7\), the seventh term is \(\frac{(-1)^{7-1}}{7!} = \frac{1}{5040} \approx 0.0002\). Thus, \(S_7 = 0.6319 + 0.0002 = 0.6321\).

Step 9: Calculate the Eighth Term

For \(n = 8\), the eighth term is \(\frac{(-1)^{8-1}}{8!} = \frac{-1}{40320} \approx -0.0000\). Thus, \(S_8 = 0.6321\).

Step 10: Analyze Convergence

After calculating the first eight partial sums: \(S_1 = 1\), \(S_2 = 0.5\), \(S_3 = 0.6667\), \(S_4 = 0.6250\), \(S_5 = 0.6333\), \(S_6 = 0.6319\), \(S_7 = 0.6321\), \(S_8 = 0.6321\). The partial sums appear to be approaching a limit.

Key Concepts

Partial Sums

In mathematics, when dealing with series, we often look at something called "partial sums." The idea behind partial sums is to add up a portion of a sequence and see where it might be heading. Imagine you have a series, which is a sequence of numbers added together. Instead of adding the whole series at once, you start by just adding up a few terms, and then a few more, and so on.

Here's how it works:

• You look at the series and decide how many terms you want to add together.
• This sum of the first few terms is called a partial sum, usually represented as \( S_n \), where \( n \) is the number of terms you're adding.
• The goal is to see if these sums are settling towards a specific number.

When the sequence of partial sums approaches a single value, we say that the series converges to that value. In the exercise above, even though each term flips sign due to the alternating series and decreases in size, you notice the partial sums getting closer to about 0.6321.

As you keep adding more terms, the changes become smaller, suggesting that the series might be convergent.

Factorials

The concept of factorials is critical in comprehending the terms of our series. When you see an exclamation mark like \( n! \) next to a number, that's a factorial. A factorial is the product of an integer and all the integers below it. For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \). Factorials grow very fast as numbers get larger. This makes them especially useful for sequences like the one in the exercise where the terms rapidly decrease.

Factorials come into play in our series formula: \( \frac{(-1)^{n-1}}{n!} \). The factorial in the denominator causes each term to shrink quite quickly, since the larger the \( n \), the larger \( n! \), and thus the smaller the fraction:

• For \( n = 1 \), you have \( \frac{1}{1} \).
• By \( n = 4 \), it becomes \( \frac{1}{24} \).
• And for \( n = 7 \), it's down to \( \frac{1}{5040} \).

The factorial ensures each following term gets smaller, contributing to the potential convergence of the series.

Alternating Series

An alternating series is a series whose terms alternate in sign. This means that every other term is positive and then negative and so on, as seen in the given series. When this pattern occurs, the series takes on the form \( a_1 - a_2 + a_3 - a_4 + \ldots \), where the \( a_n \) are positive terms.

One key result concerning alternating series is the "Alternating Series Test." It states that if the absolute value of the terms \( a_n \) in the series is decreasing and approaching zero, then the series is convergent.
If we revisit our example, \( \sum_{n=1}^{\u221E} \frac{(-1)^{n-1}}{n!} \):

• The negative sign \((-1)^{n-1}\) ensures the terms alternate. For example, when \(n=1\), term is positive; when \(n=2\), it's negative.
• Each successive term, due to the factorial in the denominator, indeed decreases in absolute value and approaches zero.

Therefore, this series with alternating signs and decreasing terms very much hints towards convergence, reflecting the behavior we've noticed in the partial sums.