Question

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

Answer 1

e-1

Answer 2

The problem asks for the sum of the infinite series $\sum_{n=1}^{\infty}\frac{1}{n!}$.

We know the Maclaurin series expansion for $e^x$ is given by: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ Substitute $x=1$ into this series: $e^1 = e = \sum_{n=0}^{\infty} \frac{1^n}{n!} = \sum_{n=0}^{\infty} \frac{1}{n!}$ Expanding the sum for $e$: $e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ Since $0! = 1$, we can write: $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ The given series is $\sum_{n=1}^{\infty}\frac{1}{n!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$. Comparing this with the expansion of $e$, we can see that: $e = 1 + \left( \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \right)$ $e = 1 + \sum_{n=1}^{\infty}\frac{1}{n!}$ To find the value of the given series, we rearrange the equation: $\sum_{n=1}^{\infty}\frac{1}{n!} = e - 1$