How to prove $ \lim_{n \to \infty} e^n \cdot \left( \sum_{k=0}^{n-1} ({k-n \over e})^k/k! \right)- 2 \cdot n = \frac 23$? I observed for the function 
$$ f(n)= e^n \sum_{k=0}^{n-1}\left(\dfrac{k - n}{e}\right)^k \cdot \dfrac{1}{k!} \tag 1$$
with small $n$ that
  n  sum
 -------------
  1  2.7182818
  2  4.6707743
  3  6.6665656
  4  8.6666045
  5  10.666662
  6  12.666667
  7  14.666667
  8  16.666667

So an obvious hypothesis is
$$ \lim_{n \to \infty} \bigl(f(n)-2n\bigr) = \frac 23 \tag 2$$
However, I have no idea, how to prove this but would like to understand how I can approach such a proof (I'll have then some similar ones with likely the same or related logic)
So I would like to understand ... 

Q: how I could prove that assumed limit (2).

 A: One can notice that your formula is the expected number of $[0,1]$-uniformly distributed random variables that are needed for their sum to exceed $n$. (See http://mathworld.wolfram.com/UniformSumDistribution.html.) Moreover, the question https://mathoverflow.net/questions/141368/error-term-for-renewal-function discusses the behaviour of the error term $\epsilon(n) = f(n) - 2n - 2/3$.
A: This is a pretty neat problem actually so I'm only giving you a little hint. Start by rewriting as
$$\sum_{k=0}^{n-1}\frac{1}{k!}(k-n)^{k}e^{-(k-n)}=\sum_{k=0}^{n-1}\frac{1}{k!}\frac{d^{k}}{dx^{k}}e^{x(k-n)}|_{x=-1}=\sum_{k=0}^{n-1}\frac{1}{2\pi i}\int_{C}\frac{e^{z(k-n)}}{(z+1)^{k+1}}dz$$
Since $e^{z(k-n)}$ is analytic in all of $\mathbb{C}$ I applied Cauchys Integral formula and $C$ denotes an appropriate cirle that encloses the multiple singularity $z=-1$
Now feel free to expand further to a geometric sum and you shouldn't be too far from a final solution. 
A: Notice that $$f'(n) = f(n) - f(n-1)$$
so this is a simple (but special) case of a delay differential equation. 
Simply substituting $f(n) = e^{an}$ reveals the form of the general solution: 
$$\sum_{-\infty}^{\infty} c_k e^{(1 + W_k(-\frac{1}{e}))n}$$ 
where $W_k(x)$ is the k-th branch of the LambertW function, and $c_k$ are parameters depending on the boundary conditions. 
To get the solution we must take into account the double root at $k=-1,0$, where  $W_{-1}(-\frac{1}{e}) = W_0(-\frac{1}{e}) = -1$:
$$f(n) = (c_{-1} +c_0n)e^0 + \sum_{-\infty,\neq 0,-1}^{\infty} c_k e^{(1 + W_k(-\frac{1}{e}))n}$$ 
but from definition $$e^{1+W_k(-\frac{1}{e})} = -\frac{1}{W_k(-\frac{1}{e})}$$ and $|\!|W_k(-\frac{1}{e})|\!| > 1$ for $k\neq, 0, -1$, so
$$\lim_{n\to\infty} f(n) = c_{-1} + c_0n + 0$$
See here for a solution that uses Laplace transform.
