Question

The sum of the following infinite series is: \[ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots \]

• A. \( \pi \)
• B. \( 1 + e \)
• C. \( e - 1 \)
• D. \( e \)

Hint:

The series for \( e^x \) is used extensively in mathematics, especially in calculus. For \( x = 1 \), the series becomes the sum for \( e - 1 \), excluding the first term.

Answer 1

The Correct Option is C

The given infinite series is:

\[\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\cdots\]

We recognize this as the expansion of the exponential constant \(e\), which is given by:

\[e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\]

From this expansion, we see that removing the first term \(1\) gives us:

\[e-1=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\]

Thus, the sum of the given series is \(e-1\).

Answer 2

Series Identification: This series is the well-known expansion of the exponential function \( e^x \) at \( x = 1 \). 
The general Taylor series expansion for \( e^x \) is: \[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \] Applying \( x = 1 \): \[ e^1 = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \] Analyzing the Given Series: The given series starts from \( \frac{1}{1!} \) instead of 1, meaning it omits the first term of the full expansion. 
Therefore, the sum of the series is: \[ \sum_{n=1}^{\infty} \frac{1}{n!} = e - 1 \] 
Conclusion: 
- This series represents \( e \) minus the first term (which is 1).
 - Hence, the sum of the series is \( e - 1 \).