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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?

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    $\begingroup$ Those are two different questions. $\endgroup$ – Angina Seng Mar 10 '18 at 18:52
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    $\begingroup$ I don't see how the first one relates to the second one. $\endgroup$ – Simply Beautiful Art Mar 10 '18 at 18:54
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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.

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