How can I prove that this sum is invariant?

Question

I found an interesting sequence that seems to give $f(n)=n!2^n$. I wasn't able to find anything online about this, so I figured I'd try to prove this. This was a verbal sequence so I turned it into a mathematical expression. After some cancelling out, the expression left was: $$f(n) = \sum_{x=0}^n(-1)^x\frac{(n-x+\frac12)^n}{x!(n-x)!}$$ Now, it seems that this function doesn't change at all, in fact it is always 1. If I had been able to prove this, I would have proven that the sequence I started with really gave $f(n)=n!2^n$. But I wasn't and that's why I came here. I know this is too specific of a question, so if it isn't suitable for here, please let me know where I can ask it.

Answer 1

Note that with a change of variables,

$$n!f(n)=\sum_{k=0}^n{\binom{n}{k}(-1)^{n-k}(k+0.5)^n}$$.

Note that $P(X)=(X+0.5)^n$ can be written as $P(X)=n!\binom{X}{n}+\sum_{0 \leq k < n}{\alpha_k\binom{X}{k}}$ for some real numbers $\alpha_k$.

Note now that if $0 \leq l < n$, then

\begin{align*} g(l) & :=\sum_{k=0}^n{(-1)^{n-k}\binom{n}{k}\binom{k}{l}}=\sum_{k=l}^n{(-1)^{n-k}\binom{n}{l}\binom{n-l}{k-l}}\\ &=\binom{n}{l}\sum_{k=0}^{n-l}{(-1)^{(n-l)-k}1^k\binom{n-l}{k}}\\ &=\binom{n}{l}(1-1)^{n-l}=0. \end{align*}

It is easy to see that $g(n)=1$.

So $$n!f(n)=\sum_{k=0}^n{(-1)^{n-k}\binom{n}{k}P(k)}=\sum_{k=0}^n{(-1)^{n-k}\binom{n}{k}\sum_{l=0}^{n-1}{\alpha_l\binom{k}{l}}+n!\binom{k}{n}}=\sum_{l=0}^{n-1}{\alpha_l\sum_{k=0}^n{(-1)^{n-k}\binom{n}{k}\binom{k}{l}}}+n!\sum_{k=0}^n{(-1)^{n-k}\binom{n}{k}\binom{k}{n}}=n!g(n)+\sum_{l=0}^{n-1}{\alpha_lg(l)}$$.

Thus, $n!f(n)=n!$ ie $f(n)=1$.