Question

Find the sum of the series: \[ 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \]

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

Hint:

The series for \( e^x \) is a well-known expansion used in calculus. For \( x = 1 \), it simplifies to the series for \( e \), which is the sum of the terms \( 1 + \frac{1}{1!} + \frac{1}{2!} + \cdots \).

Answer 1

The Correct Option is C

The given series is the standard Maclaurin series expansion for the exponential function \( e^x \), where \( x = 1 \). The general form of the series for \( e^x \) is: \[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \] Substituting \( x = 1 \): \[ e^1 = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \] Therefore, the sum of the series is \( e \), which gives the correct answer: \[ \boxed{e} \]