Infinite Series Sum Calculation

The question asks for the sum of the infinite series:

$ S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $

Using Taylor Expansion of '$e^x$'

Recall the Taylor series expansion for the exponential function '$e^x$' around $x=0$:

$ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... $

Evaluating for '$e$'

Let $x=1$. The series becomes:

$ e^1 = e = \sum_{n=0}^{\infty} \frac{1^n}{n!} $

Since $1^n = 1$ for all $n$ and $0! = 1$, we have:

$ e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... $

$ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ... $

Determining the Series Sum

The given series $ S $ is:

$ S = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ... $

Comparing the expansion of '$e$' with the series '$S$', we see:

$ e = 1 + S $

Rearranging the equation to solve for $S$:

$ S = e - 1 $

Therefore, the sum of the infinite series is $ e - 1 $.