Sum of the factorial series

Question

Is it possible somehow calculate the following sequence?

$$1 + {N!\over (N-1)!} + {N!\over (N-2)!} + {N!\over (N-3)!} +\cdot\cdot\cdot + {N!\over (3)!} + {N!\over (2)!} + {N!\over (1)!} + {N!\over (0)!}$$

Basically I don't know if it possible to simplify the following sequence:

$$1 + N! * ({1\over (N-1)!} + {1\over (N-2)!} + {1\over (N-3)!} +\cdot\cdot\cdot + {1\over (3)!} + {1\over (2)!} + {1\over (1)!} + {1\over (0)!})$$

Do we have some formula to calculate this?

Answer 1

What you have is basically

$$\sum_{k=0}^N\frac{N!}{k!}=N!\sum_{k=0}^N\frac1{k!}$$

This happens to be known as the exponential sum function where we have

$$e_N(x)=\sum_{k=0}^N\frac{x^k}{k!}$$

Yours being the case of $x=1$. It doesn't really simplify from there.