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 \(e-1\).