Key Concepts and Formulas

• Maclaurin Series for $e^x$: The Maclaurin series expansion of $e^x$ is given by: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dotsb = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ This series converges for all real and complex values of $x$.

Step-by-Step Solution

Step 1: Analyze the Given Series We are asked to find the sum of the infinite series: $S = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dotsb$ We observe that the series involves factorials in the denominator, alternating signs, and it begins with the term $\frac{1}{2!}$.

Step 2: Relate the Series to the Maclaurin Expansion of $e^x$ The Maclaurin series for $e^x$ is a powerful tool for evaluating sums of series that resemble it. The general form is: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dotsb$ We need to find a value of $x$ such that when substituted into the $e^x$ series, it produces a series similar to $S$. The alternating signs in $S$ suggest that $x$ might be negative.

Step 3: Substitute $x = -1$ into the Maclaurin Series for $e^x$ Let's substitute $x = -1$ into the Maclaurin series for $e^x$: $e^{-1} = 1 + (-1) + \frac{(-1)^2}{2!} + \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} + \frac{(-1)^5}{5!} + \dotsb$

Step 4: Simplify the Expanded Series for $e^{-1}$ Now, we simplify each term in the expansion of $e^{-1}$:

• The first term is $1$.
• The second term is $(-1) = -1$.
• The third term is $\frac{(-1)^2}{2!} = \frac{1}{2!}$.
• The fourth term is $\frac{(-1)^3}{3!} = \frac{-1}{3!} = -\frac{1}{3!}$.
• The fifth term is $\frac{(-1)^4}{4!} = \frac{1}{4!}$.
• The sixth term is $\frac{(-1)^5}{5!} = \frac{-1}{5!} = -\frac{1}{5!}$. And so on.

Substituting these back into the expansion, we get: $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dotsb$

Step 5: Isolate the Target Series Observe that the first two terms in the expansion of $e^{-1}$ are $1 - 1$, which sum to $0$. $e^{-1} = (1 - 1) + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dotsb$ $e^{-1} = 0 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dotsb$ $e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dotsb$ This is exactly the series $S$ that we were asked to evaluate.

Step 6: Determine the Sum Therefore, the sum of the given series is $e^{-1}$.

Common Mistakes & Tips

• Incorrect Substitution: Ensure you correctly substitute the chosen value of $x$ and pay close attention to the signs of the terms, especially when $x$ is negative.
• Starting Term: Be mindful of which term the given series starts with. If the standard series has initial terms that are not present in the given series, you may need to adjust by adding or subtracting those terms.
• Memorization: Having the Maclaurin series for $e^x$ memorized is crucial for quick problem-solving.

Summary

The problem requires recognizing that the given series is a modification of the Maclaurin series expansion of $e^x$. By substituting $x = -1$ into the Maclaurin series for $e^x$, we obtain $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dotsb$. After canceling the initial $1-1$ terms, we are left with the desired series, confirming that its sum is $e^{-1}$.

The final answer is $\boxed{e^{ - 1}}$.