what is the sum of following permutation series $nP0 + nP1 + nP2 +\cdots+ nPn$? what is the sum of following permutation series $nP0 + nP1 + nP2+\cdots+ nPn$ ? 
I know that
$nC0 + nC1 +\cdots + nCn = 2^n$, but not for permutation. Is there some standard result for this ?
 A: Let us call the sum $S(n)$. We have $S(1) = 1P0 + 1P1 = 1+1 = 2$.
For $n\gt 1$, we can factor out $n$ to get
$$\begin{align*} 
S(n) &= nP0 + nP1 + nP2 + \cdots + nPn\\
&= 1+ n + n(n-1) + \cdots + n!\\
&= 1+ n\Bigl( 1+(n-1) + (n-1)(n-2) + \cdots + (n-1)!\Bigr)\\
&= 1+nS(n-1).
\end{align*}$$
Thus, the sequence begins:
$$\begin{align*}
S(1) &= 2\\
S(2) &= 1 + 2S(1) = 5\\
S(3) &= 1+ 3S(2) = 16\\
S(4) &= 1+4S(3) = 65\\
S(5) &= 1+5S(4) = 326\\
&\vdots
\end{align*}$$
This is sequence A00522 on the On-Line Encyclopedia of Integer Sequences. The entry contains numerous connections; e.g., $S(n)$ is the permanent of the $n\times n$ matrix with $2$s in the diagonal and $1$s elsewhere; it also gives the formula from Marvis's post:
$$S(n) = \exp(1)\Gamma(n+1,1)\text{ where }\Gamma(z,t)=\int_{x\geq t} \exp(-x)x^{z-1}\, dx$$
A: There is no "nice" closed form as such though it is related to the incomplete $\Gamma$ function.
$P(n,k) = \dfrac{n!}{(n-k)!}$. Hence, $$\sum_{k=0}^{n} P(n,k) = n! \left( \sum_{k=0}^{n} \dfrac1{(n-k)!} \right) = n! \left( \sum_{k=0}^{n} \dfrac1{k!} \right)$$
The above sum is related to the incomplete $\Gamma$ function, which is defined as $$\Gamma(n+1,x) = \int_x^{\infty} t^{n} e^{-t} dt = n! e^{-x} \sum_{k=0}^n \left(\dfrac{x^k}{k!} \right)$$ taking $n$ to be a positive integer. Setting $x=1$, we get that
$$\Gamma(n+1,1) = \dfrac{n!}e \left(\sum_{k=0}^n \dfrac1{k!} \right)$$
Hence, $$\sum_{k=0}^{n} P(n,k) = e \times \Gamma(n+1,1)$$
A: You can write this as
$$ S(n) = n! \left( {1 \over 0!} + {1 \over 1!} + \cdots + {1 \over n!} \right) $$
and now recall that $e = 1/0! + 1/1! + 1/2! + \cdots $. So in fact 
$$ S(n) = n! \left( e - \left( {1 \over (n+1)!} + {1 \over (n+2)!} + \cdots \right) \right) $$
and we can rearrange this to give
$$ S(n) = n! \times e - \left( {1 \over (n+1)} + {1 \over (n+1)(n+2)} + \cdots \right). $$
Call the expression in parentheses $g(n)$ -- that is,
$$ g(n) = {1 \over (n+1)} + {1 \over (n+1)(n+2)} + {1 \over (n+1)(n+2)(n+3)} + \cdots $$.
Then clearly
$$ g(n) < {1 \over n} + {1 \over n^2} + {1 \over n^3} + \cdots $$
and by the usual formula for the sum of a geometric series,
$$ g(n) < {1/n \over 1-(1/n)} = {1 \over n-1}. $$
In particular if $n > 2$ we have $g(n) < 1$ and therefore $(n! \times e) - 1 < S(n) < n! \times e$.  But $S(n)$ is a sum of integers and is therefore an integer. So $S(n) = \lfloor n! \times e \rfloor$ -- that is, $S(n)$ is the greatest integer less than $n! \times e$. For example $4! \times e \approx 65.23$ and $S(4) = 65$.
This sequence is A000522 in the OEIS and the formula I gave here is given there without proof. 
Also, the number of derangements of n elements is given by $n! (1/0! - 1/1! + 1/2! - \cdots \pm 1/n!)$ and can similarly be proven to be the integer closest to $n!/e$.
